Kinematics of motion in a straight line
This unit describes motion without asking what causes it. For a body on a straight line, displacement, velocity and acceleration are linked by calculus: velocity is the derivative of displacement and acceleration is the derivative of velocity, while integration goes back the other way. The same links appear on graphs as slopes and areas. The unit ends with vector valued functions , which describe motion along curves in a plane or in space and give velocity, speed and acceleration vectors by differentiation.
Learning outcomes (12)
- Recognize distance and speed as scalar quantities, and displacement, velocity and acceleration as vector quantities.
- Sketch and interpret displacement-time and velocity-time graphs.
- Use differentiation and integration with respect to time to solve simple problems concerning displacement, velocity and acceleration.
- Use the appropriate formula for motion with constant acceleration in a straight line.
- Apply the concepts of mechanics to real life problems such as the motion of vehicles on roads, projectile motion, free fall motion and relative motion.
- Explain the need for a vector valued function.
- Construct a vector valued function.
- Identify the domain and range of vector valued functions.
- Identify the difference between scalar and vector valued functions.
- Explain the derivative of a vector function of a single variable and elaborate the result.
- Apply vector differentiation to calculate the velocity and acceleration of a position vector.
- Apply the concepts of vector valued functions to real life word problems (such as engineering and transportation).
5.1Scalar and vector quantitiesOften examined
Motion is a change in the position of a body with time. Whether a body is moving depends on the observer. A book on a table is at rest for you, but it moves with the Earth around the Sun, because you and the book share the Earth's motion. Kinematics describes motion (position, velocity, acceleration and time) without asking what force causes it.
In this unit the body moves along a straight line. We fix a point O on the line (the origin) and choose one direction as positive. Then the position of the body is one signed number: its distance from O, with a plus sign on one side of O and a minus sign on the other.
Physical quantities are of two kinds.
- Scalar quantity
- A quantity that is fully described by its magnitude (a number with a unit). Examples: time, distance, speed, mass, work, charge.
- Vector quantity
- A quantity that needs both magnitude and direction. Examples: displacement, velocity, acceleration, force, weight, momentum.
Distance (5.1.1)
Distance is the total length of the path covered by a moving body, whatever the direction of motion. It is a scalar, it is never negative, and its SI unit is the metre (). If you walk due north and then due west, the distance covered is .
Speed (5.1.2)
Speed is the rate at which distance is covered. It tells how fast a body moves, but not in which direction. It is a scalar with SI unit . A car at has the same speed whether it drives north or south.
Displacement (5.1.3)
Displacement is the change in position of a body: the straight-line distance from the initial position to the final position, together with the direction from the first to the second. It is a vector. In Figure 1 the body moves along the curved path from A to B, but its displacement is the straight arrow .
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For motion on a straight line, displacement is a signed number . A positive means the body is on the positive side of O, a negative means it is on the other side. The sign does the job of the direction.
Velocity (5.1.4)
Velocity is the rate of change of displacement with respect to time. It tells both how fast and in which direction a body moves. A car moving at due west has velocity westward. Velocity is a vector with SI unit . On a straight line, a positive velocity means motion in the positive direction and a negative velocity means motion in the negative direction.
and are the positions at times and .
Acceleration
Acceleration is the rate of change of velocity with respect to time. It is a vector with SI unit . When the acceleration is opposite to the velocity, the body slows down; this is called retardation or deceleration. A body has uniform acceleration when its velocity changes by equal amounts in equal intervals of time.
| Distance | Displacement |
|---|---|
| Scalar (magnitude only) | Vector (magnitude and direction) |
| Length of the actual path followed | Straight line from the initial to the final position |
| Never negative | Positive, negative or zero |
| Depends on the path | Depends only on the end points |
| Round trip: equals twice the one-way length | Round trip: zero |
| Speed | Velocity |
|---|---|
| Scalar: rate of change of distance | Vector: rate of change of displacement |
| Never negative | Its sign on a line shows the direction |
| Average speed = total distance ÷ total time | Average velocity = displacement ÷ time |
5.2Displacement-time graphOften examined
A displacement-time graph (s-t graph) shows where a body is at every instant. Time is the independent quantity, so it is taken along the horizontal axis (-axis). Displacement depends on time, so it is taken along the vertical axis (-axis).
Displacement-time graph for uniform motion (5.2.1)
A body is in uniform motion when it covers equal displacements in equal intervals of time, that is, when its velocity is constant. Its s-t graph is a straight line with a non-zero slope. A horizontal line (zero slope) means the displacement does not change: the body is at rest.
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Slope of a displacement-time graph (5.2.2)
Derivation Slope of the s-t graph is the velocity
Aim. Show that the slope of the displacement-time graph of a uniformly moving body equals its velocity.
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- The body starts at O and moves with uniform velocity. After time it is at P, where its displacement is . In the figure, and .
- The slope of a line is rise divided by run. For the line OP the rise is and the run is .
- By definition, displacement divided by the time taken is the velocity of the body.
- The two right-hand sides are equal, so the slope and the velocity are equal.
Result
A steeper line means a larger velocity. A line sloping down means a negative velocity. For a curved graph, the slope of the tangent at a point gives the velocity at that instant.
Time is never negative, but displacement can be positive or negative, so the slope can be positive or negative. A greater slope (in size) means a greater speed. A zero slope (horizontal line) means zero velocity.
Direction of slope (5.2.3)
- Positive slope (the line rises): displacement increases, so the body moves in the positive direction.
- Negative slope (the line falls): displacement decreases, so the body moves in the negative direction, back towards O or past it.
- Zero slope (horizontal line): the body is at rest.
- The steeper the line, the faster the body moves.
Non-uniform motion: curved s-t graphs
When the velocity changes, the s-t graph is a curve. The velocity at an instant is the slope of the tangent to the curve at that instant. The average velocity over a stage is the slope of the chord joining the two end points of the stage.
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Reading an s-t graph
A walker's s-t graph joins the points , , and , with in minutes and in metres. Find the velocity for each stage, the average velocity and the average speed for the whole walk.
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- Stage OA. Velocity is the slope of the line.
- Stage AB. The line is horizontal, so the walker is at rest.
- Stage BC. The line falls, so the velocity is negative (the walker returns).
- Whole walk. The walker ends where he started, so the displacement is zero, but the distance is .
5.2.4Velocity-time graphOften examined
A velocity-time graph (v-t graph) plots the velocity of a body against time: velocity on the -axis, time on the -axis. From one v-t graph we can read the velocity and speed at any time, the acceleration, the displacement and distance covered, and the average speed and average velocity.
- Constant velocity: the graph is a horizontal line, because the velocity is the same at every time.
- Uniform acceleration: the velocity changes by equal amounts in equal times, so the graph is a straight line that is neither horizontal nor vertical. It rises for a positive acceleration and falls for a negative acceleration (retardation).
- Variable acceleration: the graph is a curve.
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Slope of a velocity-time graph (5.2.5)
Case I: velocity is constant. The graph is a horizontal line, so its slope is zero. The acceleration is zero.
Case II: acceleration is uniform. The graph is a sloping straight line.
Derivation Slope of the v-t graph is the acceleration
Aim. Show that the slope of the velocity-time graph of a uniformly accelerated body equals its acceleration.
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- At time the velocity is (point A). At time the velocity is (point B). Draw AC parallel to the time axis and BC parallel to the velocity axis, so ACB is a right-angled triangle.
- Read the sides of the triangle from the axes: the run is the time taken and the rise is the change in velocity.
- The slope of AB is rise divided by run.
- Change in velocity divided by the time taken is, by definition, the acceleration.
Result
A rising line gives a positive acceleration, a falling line gives a negative acceleration (retardation), and a steeper line gives a larger acceleration. For a curved v-t graph, the slope of the tangent gives the acceleration at that instant.
Area between the graph and the time axis
The area between a v-t graph and the time axis over an interval gives the displacement in that interval. For constant velocity the region is a rectangle, and its area is exactly displacement = velocity × time. For uniform acceleration from to the region is a trapezium, and its area is the average velocity times the time.
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Area below the time axis (where ) counts as negative displacement. To find the distance, add all the areas as positive numbers.
Example 1
Draw the velocity-time graph of a moving particle with the given data. Find its acceleration and retardation. Also find its displacement in the time interval to .
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- Plot the points and join them with straight lines. From A to C the velocity increases, from C to E it is constant, and from E to G it decreases.
- Acceleration (A to C). The graph is a straight line, so use any two points on it, for example B(1, 5) and C(2, 8).
- C to E. The line is horizontal, so the particle moves with constant velocity and zero acceleration.
- Retardation (E to G). Use E(4, 8) and G(6, 0).The acceleration is , so the retardation is .
- Displacement from 2 s to 5 s is the area of the shaded region PCEFQ. Split it into a rectangle (2 s to 4 s) and a trapezium (4 s to 5 s).
- The book splits the same region into two rectangles and a triangle: . Both ways give the same area.
Reading s-t, v-t and a-t graphs together
An acceleration-time graph (a-t graph) plots acceleration against time. For uniform acceleration it is a horizontal line. Since velocity is the integral of acceleration, the area under an a-t graph gives the change in velocity. If the body starts from rest, that area is the velocity itself.
| Motion | s-t graph | v-t graph | a-t graph |
|---|---|---|---|
| At rest | horizontal line | on the t-axis (v=0) | on the t-axis (a=0) |
| Uniform velocity | sloping straight line | horizontal line | on the t-axis (a=0) |
| Uniform acceleration from rest | parabola curving upward | rising straight line | horizontal line above the t-axis |
| Uniform retardation | curve that flattens out | falling straight line | horizontal line below the t-axis |
One motion, three graphs
A car starts from rest and moves with uniform acceleration 2\,\mathrm{m\,s^{-2}} for 5\,\mathrm{s}. Draw its a-t, v-t and s-t graphs and check the area rules.
- Integrate the constant acceleration twice. The car starts from rest at O, so both constants are zero.a=2,\qquad v=\int 2\,dt=2t,\qquad s=\int 2t\,dt=t^2
a-t graph
v-t graph
s-t graph
v(5)=10\,\mathrm{m\,s^{-1}},\qquad s(5)=25\,\mathrm{m}
5.2Displacement function
A displacement function (position function) S(t) gives the position of a moving particle at any time t. For example, S(t)=3t^2+6t+1, with S in metres and t in seconds.
- Position at the start: put t=0.S(0)=0+0+1=1\,\mathrm{m}
- Position at t=3.S(3)=3(3)^2+6(3)+1=27+18+1=46\,\mathrm{m}
- Displacement during the first 3 seconds is final position minus initial position.S(3)-S(0)=46-1=45\,\mathrm{m}
- Average velocity over the same interval.\bar v=\dfrac{S(3)-S(0)}{3-0}=\dfrac{45}{3}=15\,\mathrm{m\,s^{-1}}
5.3Velocity as derivative of displacement functionOften examined
Average velocity over an interval is the slope of a chord of the s-t graph. To get the velocity at one instant, shrink the interval to zero. The chord turns into the tangent, and its slope becomes the derivative.
Derivation Instantaneous velocity is a derivative
Aim. Show that the velocity of a particle at time t is v=\dfrac{dS}{dt}.
- Between times t and t+\Delta t the particle moves from S(t) to S(t+\Delta t). Average velocity is displacement divided by time.\bar v=\dfrac{S(t+\Delta t)-S(t)}{\Delta t}
- On the s-t graph this is the slope of the chord joining the two points. Let \Delta t\to0: the second point slides onto the first, and the chord becomes the tangent at time t.
- The limit of the average velocity is the instantaneous velocity, and this limit is the definition of the derivative.v=\lim_{\Delta t\to0}\dfrac{S(t+\Delta t)-S(t)}{\Delta t}=\dfrac{dS}{dt}
Resultv=\dfrac{dS}{dt}
Velocity is the rate of change of displacement. Putting a particular value of t in v(t) gives the velocity at that instant. If S is in metres and t in seconds, v is in \mathrm{m\,s^{-1}}.
Example 2
A particle is moving such that its position is given by S=t^2+2\sin t. Find its velocity at any time t. Also find its initial velocity and its velocity at t=\dfrac{\pi}{3} (S in metres, t in seconds).
- Differentiate the position function with respect to t.v(t)=\dfrac{dS}{dt}=2t+2\cos t\qquad(1)
- Initial velocity: put t=0 in (1).v(0)=2(0)+2\cos0=2\,\mathrm{m\,s^{-1}}
- Put t=\dfrac{\pi}{3} in (1). Use \cos\dfrac{\pi}{3}=\dfrac12.v\!\left(\tfrac{\pi}{3}\right)=\dfrac{2\pi}{3}+2\left(\tfrac12\right)=\dfrac{2\pi}{3}+1=\dfrac{2\pi+3}{3}\approx3.09\,\mathrm{m\,s^{-1}}
v=2t+2\cos t,\quad v(0)=2\,\mathrm{m\,s^{-1}},\quad v\!\left(\tfrac{\pi}{3}\right)=\dfrac{2\pi+3}{3}\,\mathrm{m\,s^{-1}}
5.3.1Acceleration as derivative of velocity and displacementOften examined
Acceleration is the rate of change of velocity with respect to time. Since velocity is itself the derivative of displacement, acceleration is the second derivative of displacement.
a=\dfrac{dv}{dt}=\dfrac{d}{dt}\left(\dfrac{dS}{dt}\right)=\dfrac{d^2S}{dt^2}
Units: \mathrm{m\,s^{-2}} when S is in metres and t in seconds.
Example 3
The position function of a moving particle is S=\sqrt t+\ln(t+1). Find the velocity and acceleration of the particle at any instant t. Also find its velocity and acceleration at t=1 and t=4 (S in metres, t in seconds).
- Differentiate S. Write \sqrt t=t^{1/2}.v=\dfrac{dS}{dt}=\tfrac12t^{-1/2}+\dfrac{1}{t+1}=\dfrac{1}{2\sqrt t}+\dfrac{1}{t+1}\qquad(1)
- Differentiate again to get the acceleration.a=\dfrac{dv}{dt}=-\tfrac14t^{-3/2}-\dfrac{1}{(t+1)^2}\qquad(2)
- Velocity at t=1 and t=4 from (1).v(1)=\tfrac12+\tfrac12=1\,\mathrm{m\,s^{-1}},\qquad v(4)=\dfrac{1}{2(2)}+\dfrac15=\dfrac14+\dfrac15=\dfrac{9}{20}\,\mathrm{m\,s^{-1}}
- Acceleration at t=1 from (2).a(1)=-\left[\dfrac{1}{4(1)}+\dfrac{1}{(2)^2}\right]=-\left(\tfrac14+\tfrac14\right)=-\tfrac12\,\mathrm{m\,s^{-2}}
- Acceleration at t=4 from (2). Note 4^{3/2}=8, so \tfrac14\cdot4^{-3/2}=\tfrac1{32}.a(4)=-\left(\dfrac{1}{32}+\dfrac{1}{25}\right)=-\dfrac{25+32}{800}=-\dfrac{57}{800}\,\mathrm{m\,s^{-2}}
v(1)=1,\ v(4)=\tfrac{9}{20}\,\mathrm{m\,s^{-1}};\qquad a(1)=-\tfrac12,\ a(4)=-\tfrac{57}{800}\,\mathrm{m\,s^{-2}}
Rest, direction of motion and speeding upOften examined
Exercise 5.2 and the review exercise ask what a particle is doing at a given time. All the answers come from the signs of v and a.
| Condition | Meaning |
|---|---|
| v=0 | the particle is (instantaneously) at rest |
| v>0 | moving in the positive direction; S is increasing |
| v<0 | moving in the negative direction; S is decreasing |
| v changes sign at t_0 | the particle turns back at t_0 |
| a>0 / a<0 | velocity is increasing / decreasing |
| v and a have the same sign | speed is increasing (speeding up) |
| v and a have opposite signs | speed is decreasing (slowing down) |
Derivation Speeding up or slowing down
Aim. Show that the speed |v| increases when v and a have the same sign.
- Speed is the size of the velocity. Where v\ne0 we can write it with the sign of v.|v|=v\cdot\operatorname{sign}(v)
- Near a time where v\ne0 the sign of v does not change, so differentiate with the sign treated as a constant.\dfrac{d|v|}{dt}=\operatorname{sign}(v)\,\dfrac{dv}{dt}=\operatorname{sign}(v)\,a
- This is positive exactly when a has the same sign as v, that is when v\,a>0.\dfrac{d|v|}{dt}>0\iff v\,a>0
Result\text{speeding up}\iff v\,a>0,\qquad \text{slowing down}\iff v\,a<0
A negative acceleration does not always mean slowing down. A particle moving in the negative direction (v<0) with a<0 is speeding up.
Describing a motion completely
A particle moves on a line with S=t^3-6t^2+9t for t\ge0 (S in metres, t in seconds). Find when it is at rest, when it moves backwards, and when it speeds up or slows down.
- Differentiate twice.v=3t^2-12t+9=3(t-1)(t-3),\qquad a=6t-12=6(t-2)
- At rest: v=0.t=1\,\mathrm{s}\ \text{and}\ t=3\,\mathrm{s}
- Direction: v<0 between the roots of v, so the particle moves backwards for 1<t<3 and forwards for 0\le t<1 and t>3.
- Speed: compare the signs of v and a on each interval.\begin{array}{c|cccc} t & (0,1) & (1,2) & (2,3) & (3,\infty)\\ \hline v & + & - & - & + \\ a & - & - & + & + \\ \hline \text{speed} & \text{decreasing} & \text{increasing} & \text{decreasing} & \text{increasing}\end{array}
s-t graph
v-t graph
\text{rest at } t=1,3;\ \text{backwards for } 1<t<3;\ \text{speeding up on } (1,2)\cup(3,\infty)
5.3.2Displacement as an integral of velocityOften examined
Differentiation takes us from displacement to velocity. Integration takes us back.
Derivation Displacement from velocity
Aim. Express the displacement S in terms of the velocity v.
- Start from the definition of velocity.\dfrac{dS}{dt}=v
- Integrate both sides with respect to t. The left side is the integral of a derivative, which gives back S.\int\dfrac{dS}{dt}\,dt=\int v\,dt
- So S is an antiderivative of v. The constant of integration is fixed by a known position, for example S=S_0 at t=0.S=\int v\,dt
ResultS=\int v\,dt,\qquad S(t_2)-S(t_1)=\int_{t_1}^{t_2}v\,dt
The definite integral of v is the signed area under the v-t graph. It gives the displacement (change in position), not the position itself.
Displacement and total distance
When v changes sign, the particle turns back. The definite integral then subtracts the backward motion from the forward motion, so it gives the displacement. The total distance counts every part of the path as positive.
\text{displacement}=\int_{t_1}^{t_2}v\,dt,\qquad \text{distance}=\int_{t_1}^{t_2}|v|\,dt
- Find the times where v=0 inside the interval.
- Split the interval at those times.
- Find the displacement on each piece, using S(t)=\int v\,dt or definite integrals.
- Add the absolute values of the pieces to get the distance; add the pieces with their signs to get the displacement.
Displacement and distance from v(t)
A particle moves with v=t^2-4t+3\,\mathrm{m\,s^{-1}}. Find its displacement and the total distance it covers from t=0 to t=4\,\mathrm{s}.
- Factorise v to find where the particle stops.v=(t-1)(t-3)=0\ \Rightarrow\ t=1,\ t=3
- Take S(t)=\int v\,dt with S(0)=0.S(t)=\dfrac{t^3}{3}-2t^2+3t
- Positions at the end points and the turning points.S(0)=0,\quad S(1)=\tfrac43,\quad S(3)=0,\quad S(4)=\tfrac43
- Displacement is final minus initial position.S(4)-S(0)=\tfrac43\,\mathrm{m}
- Distance adds the sizes of the three pieces.\left|\tfrac43-0\right|+\left|0-\tfrac43\right|+\left|\tfrac43-0\right|=4\,\mathrm{m}
\text{displacement}=\tfrac43\,\mathrm{m},\qquad \text{distance}=4\,\mathrm{m}
5.4Velocity as an integral of accelerationOften examined
Derivation Velocity from acceleration
Aim. Express the velocity v in terms of the acceleration a.
- Start from the definition of acceleration.\dfrac{dv}{dt}=a
- Integrate both sides with respect to t.\int\dfrac{dv}{dt}\,dt=\int a\,dt
- The left side gives back v. A known velocity (for example the initial velocity) fixes the constant.v=\int a\,dt
Resultv=\int a\,dt,\qquad v(t_2)-v(t_1)=\int_{t_1}^{t_2}a\,dt
The area under the a-t graph is the change in velocity. Integrating once more gives the displacement, so from a(t) and two conditions we can find v(t) and S(t).
Example 4
The acceleration of a moving particle at any time t is a=24t+\cos t. Find its velocity and position at any time t, given that v=0 at t=0 and S=\dfrac{\pi^3}{2} at t=\dfrac{\pi}{2}.
- Integrate the acceleration; A is the constant of integration.v=\int(24t+\cos t)\,dt=12t^2+\sin t+A
- Use v=0 at t=0.0=12(0)^2+\sin0+A\ \Rightarrow\ A=0,\qquad v=12t^2+\sin t
- Integrate the velocity; B is a new constant.S=\int(12t^2+\sin t)\,dt=4t^3-\cos t+B
- Use S=\dfrac{\pi^3}{2} at t=\dfrac{\pi}{2}, with \cos\dfrac{\pi}{2}=0.\dfrac{\pi^3}{2}=4\left(\dfrac{\pi}{2}\right)^3-0+B=\dfrac{\pi^3}{2}+B\ \Rightarrow\ B=0
v=12t^2+\sin t,\qquad S=4t^3-\cos t
Equations of motion with constant accelerationOften examined
When the acceleration a is constant, integration gives three standard equations. Symbols: v_i initial velocity (at t=0), v_f final velocity (at time t), a constant acceleration, S displacement from the starting point, t time taken.
Derivation Equations of motion by calculus
Aim. Prove v_f=v_i+at, S=v_it+\tfrac12at^2 and 2aS=v_f^2-v_i^2 for constant acceleration a (Exercise 5.2, Q.13).
- Acceleration is the derivative of velocity, and a is constant.\dfrac{dv}{dt}=a
- Integrate with respect to t; C_1 is a constant.v=at+C_1
- At t=0 the velocity is v_i, so C_1=v_i. At time t the velocity is v_f.v_f=v_i+at\qquad(1)
- Velocity is the derivative of displacement.\dfrac{dS}{dt}=v_i+at
- Integrate again; C_2 is a constant. Displacement is measured from the starting point, so S=0 at t=0 and C_2=0.S=v_it+\tfrac12at^2+C_2\ \Rightarrow\ S=v_it+\tfrac12at^2\qquad(2)
- To remove t, write t from (1) and substitute in (2).t=\dfrac{v_f-v_i}{a},\qquad S=v_i\left(\dfrac{v_f-v_i}{a}\right)+\dfrac{a}{2}\left(\dfrac{v_f-v_i}{a}\right)^2
- Multiply by 2a and simplify.2aS=2v_iv_f-2v_i^2+v_f^2-2v_iv_f+v_i^2=v_f^2-v_i^2\qquad(3)
Resultv_f=v_i+at,\qquad S=v_it+\tfrac12at^2,\qquad 2aS=v_f^2-v_i^2
These three equations solve any straight-line problem with constant acceleration. They are the special case of v=\int a\,dt and S=\int v\,dt with a constant.
Derivation The same results from the v-t graph
Aim. Obtain S=v_it+\tfrac12at^2 from the area under the v-t graph.
- With constant acceleration the v-t graph is a straight line from (0,v_i) to (t,v_f). Its slope is a, so v_f-v_i=at.
- The area under the line is a rectangle of height v_i plus a triangle of height v_f-v_i.S=v_it+\tfrac12(v_f-v_i)\,t
- Replace v_f-v_i by at.S=v_it+\tfrac12(at)\,t=v_it+\tfrac12at^2
ResultS=v_it+\tfrac12at^2=\tfrac12(v_i+v_f)\,t
The area method and the calculus method agree, as they must.
Free fall and vertical motion
Near the Earth's surface, and ignoring air resistance, every body moves with the same constant acceleration g directed downwards: g=9.8\,\mathrm{m\,s^{-2}} (32\,\mathrm{ft\,s^{-2}}). The acceleration does not depend on the mass or volume of the body. Taking upwards as positive, a=-g.
v=u-gt,\qquad h=ut-\tfrac12gt^2,\qquad v^2=u^2-2gh
u: initial upward velocity, h: height above the point of projection.
5.5Application of mechanics in real life situationsOften examined
The same ideas describe vehicles on roads, falling bodies, bodies thrown upwards and the relative motion of two bodies. Choose a positive direction, write a, integrate to get v and S, and use the given conditions.
Example 5
A stone is projected vertically upwards with a velocity of 10\,\mathrm{m\,s^{-1}}. Find the height attained by the stone at any time t. When will it attain its maximum height? Also find the maximum height attained by the stone and the total distance travelled when it hits the ground.
- Take upwards as positive. The only acceleration is gravity, acting downwards.a=-g
- Integrate; A is a constant. At t=0, v=10, so A=10.v=\int -g\,dt=-gt+A\ \Rightarrow\ v=10-gt
- Integrate again; B is a constant. At t=0 the stone is at the point of projection, S=0, so B=0.S=\int(10-gt)\,dt=10t-\tfrac12gt^2+B\ \Rightarrow\ S=10t-\tfrac12gt^2
- At the highest point the stone stops for an instant, so v=0.10-gt=0\ \Rightarrow\ t=\dfrac{10}{g}\,\mathrm{s}\approx1.02\,\mathrm{s}
- Maximum height: put t=\dfrac{10}{g} in S.S=10\left(\dfrac{10}{g}\right)-\dfrac12g\left(\dfrac{10}{g}\right)^2=\dfrac{100}{g}-\dfrac{50}{g}=\dfrac{50}{g}\,\mathrm{m}\approx5.10\,\mathrm{m}
- The stone goes up \dfrac{50}{g} and comes down the same height to the ground.\text{total distance}=\dfrac{50}{g}+\dfrac{50}{g}=\dfrac{100}{g}\,\mathrm{m}\approx10.2\,\mathrm{m}
Height-time graph (g = 9.8)
Velocity-time graph
S=10t-\tfrac12gt^2,\quad t_{\max}=\dfrac{10}{g}\,\mathrm{s},\quad H=\dfrac{50}{g}\,\mathrm{m},\quad \text{distance}=\dfrac{100}{g}\,\mathrm{m}
Repeating Example 5 with a general speed u gives three results worth remembering for a body thrown up from the ground:
t_{\text{top}}=\dfrac{u}{g},\qquad H=\dfrac{u^2}{2g},\qquad t_{\text{flight}}=\dfrac{2u}{g}
Motion of vehicles on roads
Stopping distance
A car moving at 72\,\mathrm{km/h} brakes with a uniform retardation of 5\,\mathrm{m\,s^{-2}}. Find the time and distance it needs to stop.
- Convert to SI units and take the direction of motion as positive, so a=-5.v_i=72\times\dfrac{5}{18}=20\,\mathrm{m\,s^{-1}},\qquad a=-5\,\mathrm{m\,s^{-2}}
- Time to stop: v_f=0.0=20-5t\ \Rightarrow\ t=4\,\mathrm{s}
- Stopping distance: area under the v-t graph (a triangle).S=\tfrac12(20)(4)=40\,\mathrm{m}
t=4\,\mathrm{s},\qquad S=40\,\mathrm{m}
Relative motion on a straight line
When two bodies A and B move along the same line with velocities v_A and v_B, the velocity of A relative to B is how A appears to move to an observer riding on B.
v_{AB}=v_A-v_B
For example, a car at 30\,\mathrm{m\,s^{-1}} overtaking a truck at 20\,\mathrm{m\,s^{-1}} gains on it at 10\,\mathrm{m\,s^{-1}}, so it takes 5\,\mathrm{s} to gain 50\,\mathrm{m}. A train at 30\,\mathrm{m\,s^{-1}} passing a child standing still takes (length of train) ÷ 30 seconds to pass.
5.6Vector valued function
So far the position of a body was one number, because it moved on a line. A body moving in a plane needs two coordinates, and a body moving in space needs three, at every instant. A vector valued function gives all of them in one formula.
- Vector valued function
- A function whose domain is a set of real numbers (usually the time t) and whose output is a vector. In three dimensions \mathbf{r}(t)=x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat{k}, where x, y and z are real functions of t (or constants), called the component functions.
\mathbf{r}(t)=x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat{k}=\langle x(t),\,y(t),\,z(t)\rangle
The angle-bracket form \langle\ ,\ ,\ \rangle lists the components; FBISE papers use both forms.
Example: \mathbf{r}(t)=3\hat{\imath}+t\hat{\jmath}+(\sin t)\hat{k}. At t=0 the output is the vector 3\hat{\imath}; at t=\dfrac{\pi}{2} it is 3\hat{\imath}+\dfrac{\pi}{2}\hat{\jmath}+\hat{k}. The input is a number, the output is a vector.
Why we need a vector valued function
- A real function y=f(x) cannot describe a path that loops or turns back, and it cannot describe motion in space. A vector function can.
- One formula gives the position of the body at every instant, so we always know where it is and when.
- Differentiating it gives the velocity and acceleration vectors, with direction included, not just their sizes.
- It models real paths in engineering and transportation: a car on a curved road, an aircraft climbing, a projectile, a satellite, a robot arm. It is also the tool for arc length and curvature of a path.
Draw \mathbf{r}(t) from the origin O. As t changes, its tip moves and traces a curve: the path (trajectory) of the particle.
5.6.1Domain and range of a vector valued functionOften examined
The domain of \mathbf{r}(t)=x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat{k} is the set of all t for which all the component functions x(t), y(t) and z(t) are defined. If the domain is not stated, it is the intersection of the domains of the component functions. The range is the set of all output vectors \mathbf{r}(t).
| Component contains | Condition on t | Example |
|---|---|---|
| \sqrt{g(t)} (even root) | g(t)\ge0 (end point included) | \sqrt{4-t}: t\le4 |
| \ln g(t) | g(t)>0 (end point excluded) | \ln(t-1): t>1 |
| \dfrac{1}{g(t)} | g(t)\ne0 | \dfrac1{t-2}: t\ne2 |
| polynomial, e^t, a^t, \sin t, \cos t, \tan^{-1}t | all real t | t^3-4t, 2^t |
| \tan t, \sec t | t\ne\dfrac{\pi}{2}+n\pi | \sec t |
Example 6
Find the domain of \mathbf{r}(t)=\ln|t-1|\,\hat{\imath}+e^t\hat{\jmath}+\sqrt t\,\hat{k}.
- Write the components.x(t)=\ln|t-1|,\qquad y(t)=e^t,\qquad z(t)=\sqrt t
- \ln|t-1| needs |t-1|>0, that is t\ne1.D_x=(-\infty,1)\cup(1,\infty)
- e^t is defined for every real t.D_y=(-\infty,\infty)
- \sqrt t needs t\ge0.D_z=[0,\infty)
- The domain of \mathbf{r} is the intersection: start at 0 (included) and leave out 1.\begin{aligned}D&=\{(-\infty,1)\cup(1,\infty)\}\\ &\quad\cap(-\infty,\infty)\cap[0,\infty)\\ &=[0,1)\cup(1,\infty)\end{aligned}
D_{\mathbf{r}}=[0,1)\cup(1,\infty)
Another domain
Find the domain of \mathbf{r}(t)=\sqrt{t+2}\,\hat{\imath}+\dfrac{1}{t-3}\hat{\jmath}+\ln(5-t)\,\hat{k}.
- Square root: t+2\ge0.t\ge-2
- Fraction: t-3\ne0.t\ne3
- Logarithm: 5-t>0.t<5
- Take the values that satisfy all three conditions.D=[-2,3)\cup(3,5)
D=[-2,3)\cup(3,5)
Range
The range is the set of vectors the function produces. For \mathbf{r}(t)=\cos t\,\hat{\imath}+\sin t\,\hat{\jmath} every output has length \sqrt{\cos^2t+\sin^2t}=1, and every direction in the xy-plane occurs, so the range is the set of all unit vectors in the xy-plane (their tips form the unit circle). For \mathbf{r}(t)=t\hat{\imath}+t^2\hat{\jmath} the range is \{x\hat{\imath}+x^2\hat{\jmath} : x\in\mathbb R\}, whose tips lie on the parabola y=x^2.
5.6.2Construction of vector valued function
To construct a vector function for a motion, describe the motion along the x-, y- and z-axes separately, then add the three parts as x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat{k}.
Book example. A particle moves so that its motion along the x- and y-axes is circular, x=\cos t and y=\sin t, while its height along the z-axis increases linearly at 3 times the time, z=3t. Its vector function is
\mathbf{r}(t)=(\cos t)\hat{\imath}+(\sin t)\hat{\jmath}+3t\hat{k}
Building vector functions from a description
(a) A car leaves the point (1,2) (in km) and moves with constant velocity 3\hat{\imath}+4\hat{\jmath} km per hour. (b) A ball is thrown horizontally at 20\,\mathrm{m\,s^{-1}} from the top of a 45\,\mathrm{m} high cliff. Write the position vector of each at time t and find where the ball lands.
- (a) With constant velocity, each coordinate is starting value plus velocity times time.\mathbf{r}(t)=(1+3t)\hat{\imath}+(2+4t)\hat{\jmath}
- (b) Take O at the foot of the cliff, x horizontal and y upwards. Horizontally there is no acceleration; vertically the acceleration is -g=-9.8\,\mathrm{m\,s^{-2}} and the ball starts at rest vertically, 45\,\mathrm{m} up.\mathbf{r}(t)=20t\,\hat{\imath}+\left(45-4.9t^2\right)\hat{\jmath}
- The ball lands when the \hat{\jmath} component is zero.45-4.9t^2=0\ \Rightarrow\ t=\sqrt{\dfrac{45}{4.9}}\approx3.03\,\mathrm{s}
- Horizontal distance at that time.x=20(3.03)\approx60.6\,\mathrm{m}
\begin{aligned}\text{(a)}\ \ \mathbf{r}(t)&=(1+3t)\hat{\imath}+(2+4t)\hat{\jmath}\\ \text{(b)}\ \ \mathbf{r}(t)&=20t\hat{\imath}+(45-4.9t^2)\hat{\jmath}\\ &\text{lands } 60.6\,\mathrm{m}\ \text{from the cliff}\end{aligned}
5.6.3Scalar valued function
A scalar valued function gives a single real number (a scalar) for each input. From a vector valued function we get scalar valued functions by taking its magnitude |\mathbf{r}(t)| or a dot product such as \mathbf{r}(t)\cdot\mathbf{r}(t)=|\mathbf{r}(t)|^2.
Book example
Find the magnitude of \mathbf{r}(t)=3\hat{\imath}+t\hat{\jmath}+(\sin t)\hat{k} as a scalar valued function, with t\ge0 as time.
- Magnitude is the square root of the sum of the squares of the components.|\mathbf{r}(t)|=\sqrt{3^2+t^2+\sin^2t}=\sqrt{9+t^2+\sin^2t}
- The output is one number for each t, so this is a scalar valued function. Its domain is [0,\infty) because t is time.
- Its smallest value is at t=0, and it grows without limit as t increases.|\mathbf{r}(0)|=\sqrt9=3\ \Rightarrow\ \text{range}=[3,\infty)
|\mathbf{r}(t)|=\sqrt{9+t^2+\sin^2t},\qquad \text{range }[3,\infty)
| Vector valued function | Scalar valued function |
|---|---|
| Output is a vector, for example \mathbf{r}(t)=t\hat{\imath}+t^2\hat{\jmath} | Output is a real number, for example f(t)=t^2+1 |
| Has 2 or 3 components along the axes | Has one value |
| Its graph is a curve (path) traced by the tip of the vector | Its graph is a curve y=f(t) in the plane |
| Examples: position, velocity, acceleration | Examples: speed |\mathbf{v}(t)|, distance |\mathbf{r}(t)|, \mathbf{r}\cdot\mathbf{r} |
Limit and continuity of a vector valued function
Everything in calculus of vector functions is done component by component. The limit of a vector function is the vector of the limits of its components, provided each of those limits exists.
\lim_{t\to a}\mathbf{r}(t)=\left(\lim_{t\to a}x(t)\right)\hat{\imath}+\left(\lim_{t\to a}y(t)\right)\hat{\jmath}+\left(\lim_{t\to a}z(t)\right)\hat{k}
\mathbf{r} is continuous at t=a when \lim_{t\to a}\mathbf{r}(t)=\mathbf{r}(a), which happens exactly when all its components are continuous at a. Limits obey the usual rules: the limit of a sum, a scalar multiple, a dot product or a cross product is the sum, multiple, dot product or cross product of the limits.
A vector limit
Evaluate \displaystyle\lim_{t\to0}\left[\dfrac{\sin t}{t}\,\hat{\imath}+(t^2+1)\hat{\jmath}+e^{2t}\hat{k}\right].
- Take the limit of each component. The first uses the standard result \lim_{t\to0}\dfrac{\sin t}{t}=1.\lim_{t\to0}\dfrac{\sin t}{t}=1,\qquad \lim_{t\to0}(t^2+1)=1,\qquad \lim_{t\to0}e^{2t}=1
- Put the limits back as components.\lim_{t\to0}\mathbf{r}(t)=\hat{\imath}+\hat{\jmath}+\hat{k}
\hat{\imath}+\hat{\jmath}+\hat{k}
5.6.4Derivative of a vector valued functionOften examined
The derivative of a vector function is its instantaneous rate of change. If \mathbf{r}(t) is the position of an object at time t, then \mathbf{r}'(t) is its velocity at that time. \mathbf{r}(t) is differentiable at t_0 when all its components are differentiable there.
Derivation The derivative is found component by component
Aim. Show that \mathbf{r}'(t)=x'(t)\hat{\imath}+y'(t)\hat{\jmath}+z'(t)\hat{k}.
- The derivative is defined exactly as for real functions, with a vector difference on top.\mathbf{r}'(t)=\lim_{h\to0}\dfrac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}
- Subtract the vectors component by component. The unit vectors \hat{\imath},\hat{\jmath},\hat{k} are constant, so they can be taken out.\dfrac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}=\dfrac{x(t+h)-x(t)}{h}\hat{\imath}+\dfrac{y(t+h)-y(t)}{h}\hat{\jmath}+\dfrac{z(t+h)-z(t)}{h}\hat{k}
- The limit of a vector is taken component by component, and each component limit is an ordinary derivative.\mathbf{r}'(t)=x'(t)\hat{\imath}+y'(t)\hat{\jmath}+z'(t)\hat{k}
Result\mathbf{r}'(t_0)=\left.\dfrac{d\mathbf{r}}{dt}\right|_{t=t_0}=x'(t_0)\hat{\imath}+y'(t_0)\hat{\jmath}+z'(t_0)\hat{k}
Geometrically, \mathbf{r}'(t) points along the tangent to the path, in the direction of motion. Its length is the speed.
The usual rules carry over, where \mathbf{u}, \mathbf{v} are vector functions, c a constant and f a real function:
\begin{aligned}(\mathbf{u}\pm\mathbf{v})'&=\mathbf{u}'\pm\mathbf{v}', & (c\,\mathbf{u})'&=c\,\mathbf{u}', & (f\,\mathbf{u})'&=f'\,\mathbf{u}+f\,\mathbf{u}',\\ (\mathbf{u}\cdot\mathbf{v})'&=\mathbf{u}'\cdot\mathbf{v}+\mathbf{u}\cdot\mathbf{v}', & (\mathbf{u}\times\mathbf{v})'&=\mathbf{u}'\times\mathbf{v}+\mathbf{u}\times\mathbf{v}'. & & \end{aligned}
Example 7
Find the derivative of the vector function f(t)=3t^2\hat{\imath}+8t\hat{\jmath}-\dfrac{1}{t^3}\hat{k} at t=5.
- Differentiate each component. Write \dfrac1{t^3}=t^{-3}.\dfrac{df}{dt}=3\dfrac{d}{dt}(t^2)\hat{\imath}+8\dfrac{d}{dt}(t)\hat{\jmath}-\dfrac{d}{dt}(t^{-3})\hat{k}
- Use the power rule. The minus sign in front of the \hat{k} term meets the -3 from the power rule, and the two minus signs make a plus.\dfrac{df}{dt}=6t\hat{\imath}+8\hat{\jmath}-(-3t^{-4})\hat{k}=6t\hat{\imath}+8\hat{\jmath}+\dfrac{3}{t^4}\hat{k}
- Put t=5; 5^4=625.\left.\dfrac{df}{dt}\right|_{t=5}=30\hat{\imath}+8\hat{\jmath}+\dfrac{3}{625}\hat{k}
f'(5)=30\hat{\imath}+8\hat{\jmath}+\dfrac{3}{625}\hat{k}
5.6.5Velocity and acceleration of a vector valued functionOften examined
If \mathbf{r}(t) is the position vector of a particle, the first derivative gives its velocity and the second derivative gives its acceleration. The size of the velocity is the speed, and the unit vector along the velocity is the direction of motion.
\begin{aligned}\mathbf{v}(t)&=\mathbf{r}'(t)=x'(t)\hat{\imath}+y'(t)\hat{\jmath}+z'(t)\hat{k}\\ \mathbf{a}(t)&=\mathbf{v}'(t)=\mathbf{r}''(t)\\ &=x''(t)\hat{\imath}+y''(t)\hat{\jmath}+z''(t)\hat{k}\end{aligned}
\text{speed}=|\mathbf{v}(t)|=\sqrt{x'(t)^2+y'(t)^2+z'(t)^2},\qquad \text{direction}=\dfrac{\mathbf{v}(t)}{|\mathbf{v}(t)|}
Example 8
Find the velocity and acceleration of the function f(t)=2t^2\hat{\imath}+3t^4\hat{\jmath}-t^3\hat{k} at t=1.
- Differentiate each component to get the velocity.\mathbf{v}(t)=\dfrac{d}{dt}(2t^2)\hat{\imath}+\dfrac{d}{dt}(3t^4)\hat{\jmath}-\dfrac{d}{dt}(t^3)\hat{k}=4t\hat{\imath}+12t^3\hat{\jmath}-3t^2\hat{k}
- Put t=1.\mathbf{v}(1)=4\hat{\imath}+12\hat{\jmath}-3\hat{k}
- Differentiate the velocity to get the acceleration.\mathbf{a}(t)=4\hat{\imath}+36t^2\hat{\jmath}-6t\hat{k}
- Put t=1.\mathbf{a}(1)=4\hat{\imath}+36\hat{\jmath}-6\hat{k}
- Going further (often asked): the speed and direction at t=1.|\mathbf{v}(1)|=\sqrt{4^2+12^2+(-3)^2}=\sqrt{169}=13,\qquad \dfrac{\mathbf{v}(1)}{|\mathbf{v}(1)|}=\dfrac{4}{13}\hat{\imath}+\dfrac{12}{13}\hat{\jmath}-\dfrac{3}{13}\hat{k}
\mathbf{v}(1)=4\hat{\imath}+12\hat{\jmath}-3\hat{k},\qquad \mathbf{a}(1)=4\hat{\imath}+36\hat{\jmath}-6\hat{k}
Velocity and acceleration on a plane path
A particle moves with \mathbf{r}(t)=(t^2-1)\hat{\imath}+2t\hat{\jmath}. Find its velocity, speed and acceleration at t=1, and show them on the path.
- Differentiate once and twice.\mathbf{v}(t)=2t\hat{\imath}+2\hat{\jmath},\qquad \mathbf{a}(t)=2\hat{\imath}
- At t=1 the particle is at \mathbf{r}(1)=0\hat{\imath}+2\hat{\jmath}, the point (0,2).\mathbf{v}(1)=2\hat{\imath}+2\hat{\jmath},\qquad |\mathbf{v}(1)|=\sqrt{4+4}=2\sqrt2\approx2.83
- The acceleration is the same at every time.\mathbf{a}(1)=2\hat{\imath}
\mathbf{v}(1)=2\hat{\imath}+2\hat{\jmath},\quad |\mathbf{v}(1)|=2\sqrt2,\quad \mathbf{a}(1)=2\hat{\imath}
Uniform circular motionOften examined
A particle moving round a circle of radius R with constant angular speed \omega (in \mathrm{rad\,s^{-1}}) has position vector
\mathbf{r}(t)=R\cos\omega t\,\hat{\imath}+R\sin\omega t\,\hat{\jmath}
|\mathbf{r}(t)|=R at every time, so the particle stays on the circle x^2+y^2=R^2.
Derivation Velocity and acceleration in uniform circular motion
Aim. For \mathbf{r}(t)=R\cos\omega t\,\hat{\imath}+R\sin\omega t\,\hat{\jmath} find \mathbf{v} and \mathbf{a}, and show that \mathbf{v}\perp\mathbf{r} and \mathbf{a}=-\omega^2\mathbf{r}.
- Differentiate each component (chain rule: the derivative of \cos\omega t is -\omega\sin\omega t).\mathbf{v}(t)=-R\omega\sin\omega t\,\hat{\imath}+R\omega\cos\omega t\,\hat{\jmath}
- The dot product of \mathbf{r} and \mathbf{v} is zero, so the velocity is perpendicular to the radius: it is along the tangent.\mathbf{r}\cdot\mathbf{v}=-R^2\omega\sin\omega t\cos\omega t+R^2\omega\sin\omega t\cos\omega t=0
- Speed is the magnitude of \mathbf{v}; use \sin^2+\cos^2=1.|\mathbf{v}|=\sqrt{R^2\omega^2\sin^2\omega t+R^2\omega^2\cos^2\omega t}=R\omega
- Differentiate again to get the acceleration.\mathbf{a}(t)=-R\omega^2\cos\omega t\,\hat{\imath}-R\omega^2\sin\omega t\,\hat{\jmath}
- Take out -\omega^2; what is left is exactly \mathbf{r}(t).\mathbf{a}(t)=-\omega^2\left(R\cos\omega t\,\hat{\imath}+R\sin\omega t\,\hat{\jmath}\right)=-\omega^2\mathbf{r}(t)
- Since \omega^2>0, \mathbf{a} is a negative multiple of \mathbf{r}: it points opposite to the position vector, towards the centre. Its size is|\mathbf{a}|=\omega^2|\mathbf{r}|=R\omega^2=\dfrac{|\mathbf{v}|^2}{R}
Result\mathbf{v}\perp\mathbf{r},\quad |\mathbf{v}|=R\omega,\quad \mathbf{a}=-\omega^2\mathbf{r},\quad |\mathbf{a}|=R\omega^2=\dfrac{v^2}{R}
The speed is constant, yet the particle accelerates, because the direction of the velocity keeps changing. This inward (centripetal) acceleration is what a force must supply to keep a body on a circle.
The time for one revolution (period) is T=\dfrac{2\pi}{\omega}.
Circular motion with numbers
A particle moves with \mathbf{r}(t)=5\cos3t\,\hat{\imath}+5\sin3t\,\hat{\jmath} (metres, seconds). Find its speed, the size of its acceleration and its period.
- Compare with \mathbf{r}=R\cos\omega t\,\hat{\imath}+R\sin\omega t\,\hat{\jmath}.R=5\,\mathrm{m},\qquad \omega=3\,\mathrm{rad\,s^{-1}}
- Use the results just derived.|\mathbf{v}|=R\omega=15\,\mathrm{m\,s^{-1}},\qquad |\mathbf{a}|=R\omega^2=45\,\mathrm{m\,s^{-2}},\qquad T=\dfrac{2\pi}{3}\approx2.09\,\mathrm{s}
|\mathbf{v}|=15\,\mathrm{m\,s^{-1}},\quad |\mathbf{a}|=45\,\mathrm{m\,s^{-2}},\quad T=\tfrac{2\pi}{3}\,\mathrm{s}
Formula sheet
Average speed
\text{average speed}=\dfrac{\text{total distance}}{\text{total time}}
Distance counts every part of the path as positive.
Unit: \mathrm{m\,s^{-1}}
Average velocity
\bar v=\dfrac{S_2-S_1}{t_2-t_1}
S_1,S_2: positions at times t_1,t_2
Unit: \mathrm{m\,s^{-1}}
Average acceleration
\bar a=\dfrac{v_2-v_1}{t_2-t_1}
v_1,v_2: velocities at times t_1,t_2
Unit: \mathrm{m\,s^{-2}}
Slope of s-t graph
\text{slope of s-t graph}=v
The slope of the tangent gives the velocity at an instant; the slope of a chord gives the average velocity.
Unit: \mathrm{m\,s^{-1}}
Slope of v-t graph
\text{slope of v-t graph}=\dfrac{v_2-v_1}{t_2-t_1}=a
A falling line means negative acceleration (retardation).
Unit: \mathrm{m\,s^{-2}}
Area under v-t graph
\text{displacement}=\text{area under v-t graph}
Area below the time axis counts as negative displacement; add all areas as positive for distance.
Unit: \mathrm{m}
Area under a-t graph
v(t_2)-v(t_1)=\text{area under a-t graph}
Gives the change in velocity.
Unit: \mathrm{m\,s^{-1}}
Velocity as a derivative
v=\dfrac{dS}{dt}
S: displacement (position) function, t: time
Unit: \mathrm{m\,s^{-1}}
Acceleration as a derivative
a=\dfrac{dv}{dt}=\dfrac{d^2S}{dt^2}
v: velocity function
Unit: \mathrm{m\,s^{-2}}
Speeding up or slowing down
v\,a>0\Rightarrow\text{speeding up},\qquad v\,a<0\Rightarrow\text{slowing down}
Compare the signs of v and a at the given time.
Unit: none
Displacement from velocity
S=\int v\,dt,\qquad S(t_2)-S(t_1)=\int_{t_1}^{t_2}v\,dt
The constant of integration comes from a known position.
Unit: \mathrm{m}
Distance from velocity
\text{distance}=\int_{t_1}^{t_2}|v|\,dt
Split the interval where v=0 and add the sizes of the pieces.
Unit: \mathrm{m}
Velocity from acceleration
v=\int a\,dt,\qquad v(t_2)-v(t_1)=\int_{t_1}^{t_2}a\,dt
The constant comes from a known velocity.
Unit: \mathrm{m\,s^{-1}}
First equation of motion
v_f=v_i+at
v_i: initial velocity, v_f: final velocity, a: constant acceleration, t: time
Unit: \mathrm{m\,s^{-1}}
Second equation of motion
S=v_it+\tfrac12at^2
S: displacement from the starting point
Unit: \mathrm{m}
Third equation of motion
2aS=v_f^2-v_i^2
Use when the time is not given or not asked.
Unit: \mathrm{m^2\,s^{-2}} on both sides
Displacement with constant acceleration
S=\tfrac12(v_i+v_f)\,t
Area of the trapezium under the v-t graph.
Unit: \mathrm{m}
Vertical motion (upwards positive)
v=u-gt,\quad h=ut-\tfrac12gt^2,\quad v^2=u^2-2gh
u: initial upward velocity, g=9.8\,\mathrm{m\,s^{-2}}=32\,\mathrm{ft\,s^{-2}}
Units: \mathrm{m\,s^{-1}}, \mathrm{m}
Body thrown up from the ground
t_{\text{top}}=\dfrac{u}{g},\quad H=\dfrac{u^2}{2g},\quad t_{\text{flight}}=\dfrac{2u}{g}
H: maximum height
Units: \mathrm{s}, \mathrm{m}
Relative velocity on a line
v_{AB}=v_A-v_B
Velocity of A as seen from B
Unit: \mathrm{m\,s^{-1}}
Vector valued function
\mathbf r(t)=x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat{k}
x,y,z: component functions of t
Unit: as for the components
Domain of a vector function
D_{\mathbf r}=D_x\cap D_y\cap D_z
Intersection of the component domains
Unit: none
Magnitude (scalar valued)
|\mathbf r(t)|=\sqrt{x(t)^2+y(t)^2+z(t)^2}
One number for each t
Unit: as for the components
Limit of a vector function
\lim_{t\to a}\mathbf r(t)=\lim x\,\hat{\imath}+\lim y\,\hat{\jmath}+\lim z\,\hat{k}
Taken component by component
Unit: none
Derivative of a vector function
\mathbf r'(t)=x'(t)\hat{\imath}+y'(t)\hat{\jmath}+z'(t)\hat{k}
Tangent to the path
Unit: none
Velocity vector
\mathbf v(t)=\mathbf r'(t)
Unit: \mathrm{m\,s^{-1}}
Speed and direction
|\mathbf v|=\sqrt{x'^2+y'^2+z'^2},\qquad \hat{\mathbf v}=\dfrac{\mathbf v}{|\mathbf v|}
Speed is a scalar; direction is a unit vector
Unit: \mathrm{m\,s^{-1}}
Acceleration vector
\mathbf a(t)=\mathbf v'(t)=\mathbf r''(t)
Unit: \mathrm{m\,s^{-2}}
Uniform circular motion
\mathbf r=R\cos\omega t\,\hat{\imath}+R\sin\omega t\,\hat{\jmath},\quad |\mathbf v|=R\omega,\quad \mathbf a=-\omega^2\mathbf r,\quad |\mathbf a|=R\omega^2=\dfrac{v^2}{R}
R: radius, \omega: angular speed
Units: \mathrm{m\,s^{-1}}, \mathrm{m\,s^{-2}}
Period of circular motion
T=\dfrac{2\pi}{\omega}
\omega in \mathrm{rad\,s^{-1}}
Unit: \mathrm{s}
Key terms
- Motion
- A change in the position of a body with time, as seen by an observer.
- Kinematics
- The description of motion (position, velocity, acceleration, time) without considering its causes.
- Scalar quantity
- A quantity described fully by its magnitude, for example time, distance, speed, mass, work.
- Vector quantity
- A quantity that needs magnitude and direction, for example displacement, velocity, acceleration, force, momentum.
- Distance
- The total length of the path covered by a body, whatever the direction; a scalar, never negative. SI unit: \mathrm{m}.
- Displacement
- The change in position of a body: the straight line from the initial to the final position, with direction; a vector. SI unit: \mathrm{m}.
- Speed
- The rate of change of distance; how fast a body moves. A scalar. SI unit: \mathrm{m\,s^{-1}}.
- Velocity
- The rate of change of displacement with respect to time; a vector. SI unit: \mathrm{m\,s^{-1}}.
- Average speed
- Total distance divided by total time.
- Average velocity
- Displacement divided by the time taken.
- Instantaneous velocity
- The velocity at one instant: v=\dfrac{dS}{dt}, the slope of the tangent to the s-t graph.
- Acceleration
- The rate of change of velocity with respect to time, a=\dfrac{dv}{dt}. SI unit: \mathrm{m\,s^{-2}}.
- Retardation
- A negative acceleration (opposite to the velocity) that slows a body down; also called deceleration.
- Uniform motion
- Motion with constant velocity: equal displacements in equal intervals of time.
- Uniform acceleration
- Acceleration in which the velocity changes by equal amounts in equal intervals of time.
- Displacement-time graph
- A graph of displacement (vertical axis) against time (horizontal axis); its slope is the velocity.
- Velocity-time graph
- A graph of velocity against time; its slope is the acceleration and the area under it is the displacement.
- Acceleration-time graph
- A graph of acceleration against time; the area under it is the change in velocity.
- Displacement function
- A function S(t) that gives the position of a moving particle at any time t.
- Instantaneous rest
- A moment when the velocity is zero, v=0; the particle may turn back there.
- Free fall
- Motion under gravity alone, with constant acceleration g=9.8\,\mathrm{m\,s^{-2}} downwards, independent of mass.
- Relative velocity
- The velocity of one body as seen from another: v_{AB}=v_A-v_B.
- Vector valued function
- A function whose domain is a set of real numbers and whose output is a vector, \mathbf r(t)=x(t)\hat{\imath}+y(t)\hat{\jmath}+z(t)\hat{k}.
- Component functions
- The real functions x(t), y(t), z(t) that make up a vector valued function.
- Domain of a vector valued function
- The set of all t for which every component function is defined: the intersection of the component domains.
- Range of a vector valued function
- The set of all output vectors of the function.
- Scalar valued function
- A function that gives a single real number for each input, for example |\mathbf r(t)|.
- Derivative of a vector function
- \mathbf r'(t)=\lim_{h\to0}\dfrac{\mathbf r(t+h)-\mathbf r(t)}{h}, found by differentiating each component; it is tangent to the path.
- Speed (vector motion)
- The magnitude of the velocity vector, |\mathbf v(t)|=|\mathbf r'(t)|.
- Helix
- A space curve such as \mathbf r(t)=\cos t\,\hat{\imath}+\sin t\,\hat{\jmath}+3t\hat{k} that winds round an axis while rising steadily.
- Uniform circular motion
- Motion round a circle at constant speed; the acceleration -\omega^2\mathbf r points to the centre.
- Centripetal acceleration
- The inward acceleration of circular motion, of size v^2/R=R\omega^2.
Exercise 5.1p. 145–147
Read velocities as slopes of s-t graphs and displacements as areas under v-t graphs. Average speed is always total distance over total time.
A cyclist rides in a straight line for 20 minutes. He waits for half an hour, then returns in a straight line to his starting point in 15 minutes. This is a displacement-time graph for his journey.
Solution
(i) Work out the average velocity for each stage of the journey in km/h.
- Stage OA (0 to 20 min): displacement 5\,\mathrm{km} in 20\,\mathrm{min}=\tfrac13\,\mathrm{h}. Velocity is displacement divided by time.v_{OA}=\dfrac{5}{1/3}=15\,\mathrm{km/h}
- Stage AB (20 to 50 min): the graph is horizontal, the displacement does not change, so he is at rest.v_{AB}=\dfrac{5-5}{1/2}=0\,\mathrm{km/h}
- Stage BC (50 to 65 min): displacement changes from 5 to 0\,\mathrm{km} in 15\,\mathrm{min}=\tfrac14\,\mathrm{h}. The minus sign shows he rides back towards the start.v_{BC}=\dfrac{0-5}{1/4}=-20\,\mathrm{km/h}
15\,\mathrm{km/h},\quad 0\,\mathrm{km/h},\quad -20\,\mathrm{km/h}
(ii) Write down the average velocity for the whole journey.
- He finishes at his starting point, so the total displacement is zero.\bar v=\dfrac{\text{displacement}}{\text{time}}=\dfrac{0}{65/60}=0\,\mathrm{km/h}
\bar v=0\,\mathrm{km/h}
(iii) Work out the average speed for the whole journey.
- Distance counts both directions: 5\,\mathrm{km} out and 5\,\mathrm{km} back.\text{distance}=5+5=10\,\mathrm{km}
- Total time, including the wait, is 65\,\mathrm{min}=\tfrac{65}{60}\,\mathrm{h}.\text{average speed}=\dfrac{10}{65/60}=\dfrac{600}{65}=\dfrac{120}{13}\approx9.23\,\mathrm{km/h}
\text{average speed}\approx9.23\,\mathrm{km/h}
This is a displacement-time graph for a car travelling along a straight road. The journey is divided into 5 stages labelled A to E.
Solution
(i) Workout the average velocity for each stage of the journey.
| Stage | Time | Displacement | Time taken | Average velocity |
|---|---|---|---|---|
| A | 9:30 to 10:00 | 40-0=40\,\mathrm{km} | 0.5\,\mathrm{h} | 80\,\mathrm{km/h} |
| B | 10:00 to 10:30 | 60-40=20\,\mathrm{km} | 0.5\,\mathrm{h} | 40\,\mathrm{km/h} |
| C | 10:30 to 11:00 | 60-60=0 | 0.5\,\mathrm{h} | 0 (at rest) |
| D | 11:00 to 12:00 | 100-60=40\,\mathrm{km} | 1\,\mathrm{h} | 40\,\mathrm{km/h} |
| E | 12:00 to 13:30 | 0-100=-100\,\mathrm{km} | 1.5\,\mathrm{h} | -66.7\,\mathrm{km/h} |
Each velocity is the slope of that stage: displacement divided by time taken. Stage E has a negative velocity because the car drives back to its starting point.
80,\ 40,\ 0,\ 40,\ -66.7\,\mathrm{km/h}
(ii) State the average velocity for the whole journey.
- The car ends where it started (displacement 0 at 13:30), so the total displacement is zero.\bar v=\dfrac{0}{4}=0\,\mathrm{km/h}
\bar v=0\,\mathrm{km/h}
(iii) Workout the average speed for the whole journey.
- Distance: 100\,\mathrm{km} out (stages A to D) and 100\,\mathrm{km} back (stage E). Time: 9:30 to 13:30 is 4\,\mathrm{h}.\text{average speed}=\dfrac{200}{4}=50\,\mathrm{km/h}
50\,\mathrm{km/h}
Note. The book labels the time axis “Time (minutes)”, but its marks are clock times from 9:30 to 13:30. Stage A ends at point A, stage B at point B, and so on.
Fatima left home at 10:00 and cycled north-east in a straight line. The diagram shows a displacement time graph for her journey.
Solution
(i) Find the Fatima's velocity between 10:00 and 11:00.
- The graph rises from 0 to 12\,\mathrm{km} in 1\,\mathrm{h}. Velocity is the slope.v=\dfrac{12-0}{11-10}=12\,\mathrm{km/h}\ \text{(north-east)}
12\,\mathrm{km/h}
(ii) On her return journey, Fatima continued passed her home before returning. Estimate the time that Fatima passed her home.
- On the stage from Q(11:30, 12) to R(13:00, −3) the displacement falls by 15\,\mathrm{km} in 1.5\,\mathrm{h}.v=\dfrac{-3-12}{13-11.5}=-10\,\mathrm{km/h}
- She passes home when s=0. Starting from 12\,\mathrm{km} and losing 10\,\mathrm{km} each hour, she needs 1.2\,\mathrm{h} after 11:30.12-10(t-11.5)=0\ \Rightarrow\ t-11.5=1.2\,\mathrm{h}=1\,\mathrm{h}\ 12\,\mathrm{min}
- Add this to 11:30. This is where the graph crosses the time axis.11{:}30+1{:}12=12{:}42
\text{about } 12{:}42
(iii) Find Fatima's velocity for each of the last two stages for her journey.
- 11:30 to 13:00 (Q to R): she rides back through her home to 3\,\mathrm{km} on the other side.v=\dfrac{-3-12}{1.5}=-10\,\mathrm{km/h}
- 13:00 to 14:00 (R to S): she rides from -3\,\mathrm{km} back to home.v=\dfrac{0-(-3)}{1}=+3\,\mathrm{km/h}
-10\,\mathrm{km/h}\ \text{and}\ 3\,\mathrm{km/h}
(iv) Calculate Fatima's average speed for her entire journey.
- Add the distances of all stages as positive numbers (the rest from 11:00 to 11:30 adds nothing).\text{distance}=12+0+15+3=30\,\mathrm{km}
- The journey lasts from 10:00 to 14:00, that is 4\,\mathrm{h}.\text{average speed}=\dfrac{30}{4}=7.5\,\mathrm{km/h}
7.5\,\mathrm{km/h}
Note. The graph does not print a unit for displacement; kilometres are assumed, as the ride lasts hours. The corners are read from the grid as (11:00, 12), (11:30, 12), (13:00, −3) and (14:00, 0); the labels P, Q, R, S are added here for reference.
An electric train starts from the rest at a station A and moves along a straight level track. The train accelerates uniformly at 0.4\,\mathrm{m/s^2} to a speed of 16\,\mathrm{m/s}. The speed is then maintained for a displacement of 2000\,\mathrm{m}. Finally the train retards uniformly for 20\,\mathrm{s} before coming to rest at a station B. For this journey from A to B,
Solution
(i) Find the total time taken.
Given
- a_1 = 0.4\,\mathrm{m\,s^{-2}} acceleration
- v = 16\,\mathrm{m\,s^{-1}} top speed
- S_2 = 2000\,\mathrm{m} at steady speed
- t_3 = 20\,\mathrm{s} retarding
To find
- total time t
- Stage 1. Starting from rest, use v_f=v_i+at.16=0+0.4\,t_1\ \Rightarrow\ t_1=40\,\mathrm{s}
- Stage 2. At constant speed, time = distance ÷ speed.t_2=\dfrac{2000}{16}=125\,\mathrm{s}
- Stage 3 is given.t_3=20\,\mathrm{s}
- Add the three stages.t=40+125+20=185\,\mathrm{s}
v-t graph of the journey
185\,\mathrm{s}
(ii) Find the displacement from A to B.
- Displacement is the area under the v-t graph above. Stage 1 is a triangle.S_1=\tfrac12(40)(16)=320\,\mathrm{m}
- Stage 2 is given.S_2=2000\,\mathrm{m}
- Stage 3 is a triangle of base 20\,\mathrm{s} and height 16\,\mathrm{m\,s^{-1}}.S_3=\tfrac12(20)(16)=160\,\mathrm{m}
- Add the stages.AB=320+2000+160=2480\,\mathrm{m}
AB=2480\,\mathrm{m}=2.48\,\mathrm{km}
(iii) Sketch the displacement-time graph, showing clearly the shape of the graph for each stage of the journey.
- Stage 1 (0 to 40 s): the velocity increases, so the s-t graph is a curve getting steeper (a parabola, s=0.2t^2), ending at (40,320).
- Stage 2 (40 to 165 s): constant velocity, so a straight line of slope 16, from (40,320) to (165,2320).
- Stage 3 (165 to 185 s): the velocity decreases to zero, so the curve gets flatter and ends horizontal at (185,2480).
Using the following data, draw time-displacement graph for a moving object. Use the graph to find average velocity for first 4\,s, for next 4\,s and for last 6\,s and the total displacement.
| Time (s) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 |
|---|---|---|---|---|---|---|---|---|---|
| Displacement (m) | 0 | 2 | 4 | 4 | 4 | 6 | 4 | 2 | 0 |
| Time (s) | Displacement (m) |
|---|---|
| 0 | 0 |
| 2 | 2 |
| 4 | 4 |
| 6 | 4 |
| 8 | 4 |
| 10 | 6 |
| 12 | 4 |
| 14 | 2 |
| 16 | 0 |
Solution
- First 4 s (0 to 4 s): average velocity is the change in displacement over the time.\bar v=\dfrac{s(4)-s(0)}{4}=\dfrac{4-0}{4}=1\,\mathrm{m\,s^{-1}}
- Next 4 s (4 to 8 s): the graph is horizontal, the object is at rest.\bar v=\dfrac{s(8)-s(4)}{4}=\dfrac{4-4}{4}=0
- Last 6 s (10 to 16 s): the displacement falls from 6 to 0, so the object moves back.\bar v=\dfrac{s(16)-s(10)}{6}=\dfrac{0-6}{6}=-1\,\mathrm{m\,s^{-1}}
- Total displacement (0 to 16 s): the object ends at its starting position.s(16)-s(0)=0-0=0\,\mathrm{m}The total distance is 4+2+6=12\,\mathrm{m} (up to 4, still, up to 6, back to 0).
1\,\mathrm{m\,s^{-1}},\quad 0,\quad -1\,\mathrm{m\,s^{-1}};\qquad \text{total displacement}=0
Ahmed leaves home at 11 am. He cycles at a speed of 16\,km/h for 90 minutes. He stops for half an hour. Ahmed then cycles home and arrives at 3 pm.
Solution
(i) Draw a displacement-time graph to show Ahmed's journey.
- Out: 90 minutes is 1.5\,\mathrm{h}, so he goes 16\times1.5=24\,\mathrm{km} and arrives at 12:30.(11{:}00,\,0)\to(12{:}30,\,24)
- Stop: half an hour at 24\,\mathrm{km}, until 13:00. The graph is horizontal.(12{:}30,\,24)\to(13{:}00,\,24)
- Return: home (s=0) at 3 pm = 15:00.(13{:}00,\,24)\to(15{:}00,\,0)
(ii) What is Ahmed's average speed on the return part of his cycle.
- He returns 24\,\mathrm{km} between 13:00 and 15:00, that is in 2\,\mathrm{h}.\text{speed}=\dfrac{24}{2}=12\,\mathrm{km/h}
12\,\mathrm{km/h}
Dabeer leaves at 14:00. He drives at an average speed of 14\,km/h for 3\tfrac12 hours. Dabeer stops the journey for 30 minutes. He then drives home at 70\,km/h. Draw a displacement-time graph to show Dabeer's journey.
Solution
- Out: distance = speed × time.14\times3.5=49\,\mathrm{km},\ \text{reached at } 14{:}00+3{:}30=17{:}30
- Stop: 30 minutes, from 17:30 to 18:00, at 49\,\mathrm{km}.
- Return: time = distance ÷ speed.\dfrac{49}{70}=0.7\,\mathrm{h}=42\,\mathrm{min},\ \text{home at } 18{:}42
(14{:}00,0)\to(17{:}30,49)\to(18{:}00,49)\to(18{:}42,0)
A helicopter leaves Islamabad at 09:00. It flies for 45 minutes at 80\,km/h. It lands for 20 minutes. The helicopter then returns to its base in Islamabad, flying at 100\,km/h. Draw a displacement-time graph to show the journey.
Solution
- Out: 45 minutes is 0.75\,\mathrm{h}.80\times0.75=60\,\mathrm{km},\ \text{landing at } 09{:}45
- Landed: 20 minutes, from 09:45 to 10:05.
- Return: time = distance ÷ speed.\dfrac{60}{100}=0.6\,\mathrm{h}=36\,\mathrm{min},\ \text{back at } 10{:}05+0{:}36=10{:}41
(09{:}00,0)\to(09{:}45,60)\to(10{:}05,60)\to(10{:}41,0)
The diagram shows the velocity-time graph of the motion of an athlete running along a straight track. For the first 4\,s, he accelerates uniformly from rest to a velocity of 9\,m/s. This velocity is then maintained for a further 8\,s. Find:
Solution
(i) the rate at which the athlete accelerates.
- Acceleration is the slope of the first part of the v-t graph.a=\dfrac{9-0}{4-0}=2.25\,\mathrm{m\,s^{-2}}
a=2.25\,\mathrm{m\,s^{-2}}
(ii) the displacement from the starting point of the athlete after 12\,s.
- Displacement is the area under the graph: a triangle (0 to 4 s) plus a rectangle (4 to 12 s).S=\tfrac12(4)(9)+(8)(9)=18+72=90\,\mathrm{m}
90\,\mathrm{m}
The diagram shows the velocity-time of the motion of a cyclist riding along a straight road. He accelerates uniformly from rest to 8\,m/s in 20\,s. He then travels at a constant velocity of 8\,m/s for 40\,s. Then he decelerates uniformly to rest in 15\,s. Find:
Solution
(i) The acceleration of the cyclist in the first 20\,s of motion.
- Slope of the first line.a=\dfrac{8-0}{20}=0.4\,\mathrm{m\,s^{-2}}
0.4\,\mathrm{m\,s^{-2}}
(ii) The deceleration of the cyclist in the last 15\,s of motion.
- Slope of the last line is negative; its size is the deceleration.\text{slope}=\dfrac{0-8}{75-60}=-\dfrac{8}{15}\ \Rightarrow\ \text{deceleration}=\dfrac{8}{15}\approx0.53\,\mathrm{m\,s^{-2}}
\dfrac{8}{15}\approx0.53\,\mathrm{m\,s^{-2}}
(iii) The displacement from the starting point of the cyclist after 75\,s.
- Area of the trapezium under the graph: parallel sides 40\,\mathrm{s} (top) and 75\,\mathrm{s} (bottom), height 8\,\mathrm{m\,s^{-1}}.S=\tfrac12(40+75)(8)=460\,\mathrm{m}
- Check by pieces: triangle, rectangle, triangle.\tfrac12(20)(8)+40(8)+\tfrac12(15)(8)=80+320+60=460\,\mathrm{m}
460\,\mathrm{m}
A particle moves 100\,m in a straight line. The diagram is a sketch of velocity-time graph of the motion of particle. The particle starts with velocity u\,m/s and accelerates to a velocity of 10\,m/s. The velocity of 10\,m/s is maintained for 7\,s and then the particle decelerates to rest in a further 2\,s. Find:
Solution
(i) The value of u.
- The total displacement (100\,\mathrm{m}) is the area under the graph. Split it into a trapezium (0 to 3 s), a rectangle (3 to 10 s) and a triangle (10 to 12 s).\tfrac12(u+10)(3)+(10)(7)+\tfrac12(2)(10)=100
- Simplify.1.5(u+10)+70+10=100\ \Rightarrow\ 1.5(u+10)=20
- Solve for u.u+10=\dfrac{40}{3}\ \Rightarrow\ u=\dfrac{10}{3}\approx3.33\,\mathrm{m\,s^{-1}}
u=\dfrac{10}{3}\,\mathrm{m\,s^{-1}}
(ii) The acceleration of the particle in the first part of the motion.
- Slope of the first line, from u=\tfrac{10}{3} to 10 in 3\,\mathrm{s}.a=\dfrac{10-\tfrac{10}{3}}{3}=\dfrac{20/3}{3}=\dfrac{20}{9}\approx2.22\,\mathrm{m\,s^{-2}}
a=\dfrac{20}{9}\approx2.22\,\mathrm{m\,s^{-2}}
A car is moving along a straight road. When t=0\,s, the car passes a point A with velocity 10\,m/s and this velocity is maintained until t=30\,s. The driver then applies the brakes and the car decelerates uniformly, coming to rest at point B when t=42\,s.
Solution
(i) Sketch a velocity-time graph to illustrate the motion of the car.
- Constant velocity 10\,\mathrm{m\,s^{-1}} from t=0 to t=30 gives a horizontal line; uniform deceleration to rest at t=42 gives a straight line down to the time axis.
(ii) Find the distance from A to B.
- Area under the graph: rectangle (0 to 30 s) plus triangle (30 to 42 s).AB=(30)(10)+\tfrac12(12)(10)=300+60=360\,\mathrm{m}
- Or use the trapezium formula with parallel sides 30 and 42.\tfrac12(30+42)(10)=360\,\mathrm{m}
AB=360\,\mathrm{m}
A particle moves along a straight line. The particle accelerates from rest to a velocity of 20\,m/s in 15\,s. The particle then moves at a constant velocity of 20\,m/s for a period of time. The particle then decelerates uniformly to rest. The period of time for which the particle is travelling at a constant velocity is 4 times the period of time for which it decelerating.
Solution
(i) Sketch a velocity-time graph to illustrate the motion of the particle.
- Let the constant-velocity stage last T seconds. Then the deceleration lasts \tfrac{T}{4} seconds. The graph is a trapezium: up to 20 in 15 s, level for T s, down to 0 in \tfrac T4 s.
(ii) Given that the displacement from the starting point of the particle after it comes to rest is 480\,m. Find the total time for which the particle is moving.
- Area under the graph = displacement. Add the triangle, rectangle and triangle.\tfrac12(15)(20)+20T+\tfrac12\left(\tfrac T4\right)(20)=480
- Simplify.150+20T+2.5T=480\ \Rightarrow\ 22.5T=330
- Solve for T.T=\dfrac{330}{22.5}=\dfrac{44}{3}\approx14.67\,\mathrm{s}
- Total time is 15+T+\tfrac T4.15+\dfrac{44}{3}+\dfrac{11}{3}=15+\dfrac{55}{3}=\dfrac{100}{3}\approx33.3\,\mathrm{s}
\dfrac{100}{3}\,\mathrm{s}\approx33.3\,\mathrm{s}
A motorcyclist M leaves a road junction at time t=0\,s. He accelerates from rest at a rate of 3\,m/s^2 for 8\,s and then maintains the velocity he has reached. A car C leaves the same road junction as M at time t=0\,s. The car accelerates from rest to 30\,m/s in 20\,s and then maintains a velocity of 30\,m/s. C passes M as they both pass a pedestrian.
Solution
(i) On the same diagram, sketch velocity-time graphs to illustrate the motion of M and C.
- M: velocity after 8 s is 3\times8=24\,\mathrm{m\,s^{-1}}, then constant: v_M=\min(3t,\,24).
- C: acceleration \tfrac{30}{20}=1.5\,\mathrm{m\,s^{-2}} for 20 s, then 30\,\mathrm{m\,s^{-1}}: v_C=\min(1.5t,\,30).
(ii) Find the distance of the pedestrian from the road junction.
- Distance of M from the junction (area under M's graph) for t\ge8: triangle up to 8 s, then 24\,\mathrm{m\,s^{-1}}.s_M=\tfrac12(8)(24)+24(t-8)=24t-96
- Distance of C for t\ge20: triangle up to 20 s, then 30\,\mathrm{m\,s^{-1}}.s_C=\tfrac12(20)(30)+30(t-20)=30t-300
- C can only catch M after reaching its top speed (for 8\le t\le20, s_C=0.75t^2=24t-96 has no root in that interval). Set s_C=s_M.30t-300=24t-96\ \Rightarrow\ 6t=204\ \Rightarrow\ t=34\,\mathrm{s}
- Both are at the pedestrian at this moment.s=30(34)-300=720\,\mathrm{m},\qquad \text{check: } 24(34)-96=720\,\mathrm{m}
720\,\mathrm{m}\ \text{from the junction}
Exercise 5.2p. 151–153
Differentiate to go from position to velocity to acceleration; integrate (with a given condition for each constant) to go back. Check the zeros of v before finding a distance.
A projectile is launched vertically upward from an initial height of 129\,ft with an initial velocity of 87\,ft/s.
Solution
(i) What are the position, velocity, and acceleration functions?
- Gravity is the only acceleration and it acts downwards.a(t)=-32\,\mathrm{ft\,s^{-2}}
- Integrate; the constant is the initial velocity 87.v(t)=\int-32\,dt=-32t+C_1,\quad v(0)=87\Rightarrow C_1=87,\quad v(t)=87-32t
- Integrate again; the constant is the initial height 129.s(t)=\int(87-32t)\,dt=87t-16t^2+C_2,\quad s(0)=129\Rightarrow C_2=129
s(t)=129+87t-16t^2,\quad v(t)=87-32t,\quad a(t)=-32
(ii) When will the projectile hit the ground?
- At the ground the height is zero.-16t^2+87t+129=0\ \Longrightarrow\ 16t^2-87t-129=0
- Quadratic formula with a=16, b=-87, c=-129.t=\dfrac{87\pm\sqrt{87^2+4(16)(129)}}{32}=\dfrac{87\pm\sqrt{15825}}{32}=\dfrac{87\pm125.80}{32}
- Time cannot be negative, so take the plus sign.t=\dfrac{212.80}{32}\approx6.65\,\mathrm{s}
t\approx6.65\,\mathrm{s}
(iii) What is its impact velocity?
- Put the landing time in v(t). Exactly, v=87-(87+\sqrt{15825})=-\sqrt{15825}.v=87-32(6.65)\approx-125.8\,\mathrm{ft\,s^{-1}}
- The minus sign means the projectile is moving downwards when it hits the ground.
v\approx-125.8\,\mathrm{ft\,s^{-1}}\ (125.8\,\mathrm{ft\,s^{-1}}\ \text{downwards})
(iv) When will the projectile reach its maximum height?
- At the highest point the velocity is zero.87-32t=0\ \Rightarrow\ t=\dfrac{87}{32}\approx2.72\,\mathrm{s}
t=\dfrac{87}{32}\approx2.72\,\mathrm{s}
(v) What is the maximum height of projectile?
- Put t=\dfrac{87}{32} in s(t).s=129+87\left(\dfrac{87}{32}\right)-16\left(\dfrac{87}{32}\right)^2=129+\dfrac{7569}{32}-\dfrac{7569}{64}=129+\dfrac{7569}{64}
- Simplify.s=\dfrac{15825}{64}\approx247.3\,\mathrm{ft}
s_{\max}\approx247.3\,\mathrm{ft}
Note. The data are in feet, so use g=32\,\mathrm{ft\,s^{-2}}. Heights are measured from the ground, upwards positive.
An object has its position defined by S=t^3-9t^2+24t+20 in feet.
Solution
(i) What are the velocity and acceleration functions?
- Differentiate S once for velocity and twice for acceleration.v=\dfrac{dS}{dt}=3t^2-18t+24,\qquad a=\dfrac{dv}{dt}=6t-18
v=3t^2-18t+24\ \mathrm{ft\,s^{-1}},\quad a=6t-18\ \mathrm{ft\,s^{-2}}
(ii) What are the position and velocity of the object when its acceleration is -6.5\,ft/s^2?
- Set a=-6.5 and solve for t.6t-18=-6.5\ \Rightarrow\ t=\dfrac{11.5}{6}=\dfrac{23}{12}\approx1.917\,\mathrm{s}
- Position at this time.S=\left(\tfrac{23}{12}\right)^3-9\left(\tfrac{23}{12}\right)^2+24\left(\tfrac{23}{12}\right)+20=7.041-33.063+46+20\approx39.98\,\mathrm{ft}
- Velocity at this time.v=3\left(\tfrac{23}{12}\right)^2-18\left(\tfrac{23}{12}\right)+24=11.021-34.5+24\approx0.52\,\mathrm{ft\,s^{-1}}
S=\dfrac{69083}{1728}\approx39.98\,\mathrm{ft},\qquad v=\dfrac{25}{48}\approx0.52\,\mathrm{ft\,s^{-1}}
(iii) Find the displacement and the total distance travelled by the particle from t=1.5\,s to t=7\,s.
- Find where the object turns back: v=0.3t^2-18t+24=3(t-2)(t-4)=0\ \Rightarrow\ t=2,\ t=4
- Positions at the end points and at the turning points.S(1.5)=39.125,\quad S(2)=40,\quad S(4)=36,\quad S(7)=90
- Displacement is final minus initial position.S(7)-S(1.5)=90-39.125=50.875\,\mathrm{ft}
- Distance adds the sizes of the three stages.|40-39.125|+|36-40|+|90-36|=0.875+4+54=58.875\,\mathrm{ft}
\text{displacement}=50.875\,\mathrm{ft},\qquad \text{distance}=58.875\,\mathrm{ft}
A person is standing on top of the Meinar-e-Pakistan and throws a ball directly upward with an initial velocity of 96\,ft/s. The Meinar-e-Pakistan is 176\,ft high.
Solution
(i) What are the functions for position, velocity, and acceleration of the ball?
- Use g=32\,\mathrm{ft\,s^{-2}} downwards.a(t)=-32
- Integrate with v(0)=96.v(t)=96-32t
- Integrate with s(0)=176 (the ball starts at the top of the tower).s(t)=176+96t-16t^2
s(t)=176+96t-16t^2,\quad v(t)=96-32t,\quad a(t)=-32
(ii) When does the ball hit the ground and with what velocity?
- At the ground s=0. Divide by -16.176+96t-16t^2=0\ \Rightarrow\ t^2-6t-11=0
- Solve and keep the positive root.t=\dfrac{6\pm\sqrt{36+44}}{2}=3\pm2\sqrt5\ \Rightarrow\ t=3+2\sqrt5\approx7.47\,\mathrm{s}
- Velocity at that time.v=96-32(3+2\sqrt5)=-64\sqrt5\approx-143.1\,\mathrm{ft\,s^{-1}}
t=3+2\sqrt5\approx7.47\,\mathrm{s},\qquad v\approx-143.1\,\mathrm{ft\,s^{-1}}\ \text{(downwards)}
(iii) How far does the ball travel during its flight?
- The ball turns back when v=0.96-32t=0\ \Rightarrow\ t=3\,\mathrm{s},\qquad s(3)=176+288-144=320\,\mathrm{ft}
- Upward trip: from 176 ft to 320 ft. Downward trip: from 320 ft to the ground.\text{up}=320-176=144\,\mathrm{ft},\qquad \text{down}=320\,\mathrm{ft}
- Add both.\text{distance}=144+320=464\,\mathrm{ft}
464\,\mathrm{ft}
A particle moves along a line such that its position is: S=2t^3-9t^2+12t-4, for t\ge0.
Solution
(i) Find t for which the distance S is increasing.
- S increases when its derivative, the velocity, is positive.v=\dfrac{dS}{dt}=6t^2-18t+12=6(t-1)(t-2)
- The product is positive when both factors have the same sign: before 1 or after 2.v>0\iff 0\le t<1\ \text{or}\ t>2
S \text{ increases for } 0\le t<1 \text{ and } t>2
(ii) Find t for which the velocity is increasing.
- Velocity increases when its derivative, the acceleration, is positive.a=\dfrac{dv}{dt}=12t-18>0\ \Rightarrow\ t>\dfrac32
t>\dfrac32
(iii) Find t for which the speed of the particle is increasing.
- Speed increases when v and a have the same sign. The sign changes are at t=1,\ \tfrac32,\ 2.\begin{array}{c|cccc} t & (0,1) & (1,\tfrac32) & (\tfrac32,2) & (2,\infty)\\ \hline v & + & - & - & + \\ a & - & - & + & + \\ \hline \text{speed} & \downarrow & \uparrow & \downarrow & \uparrow\end{array}
- Read off the intervals where the signs agree.
1<t<\dfrac32\ \text{ and }\ t>2
(iv) Find the speed when t=\dfrac32\,s.
- Put t=\tfrac32 in v.v\left(\tfrac32\right)=6\left(\tfrac12\right)\left(-\tfrac12\right)=-\tfrac32
- Speed is the size of the velocity.\text{speed}=\left|-\tfrac32\right|=\tfrac32=1.5\ \text{units per second}
\text{speed}=1.5
(v) Find the total distance travelled in the time interval [0,4].
- The particle turns at t=1 and t=2 (where v=0). Find the positions.S(0)=-4,\quad S(1)=1,\quad S(2)=0,\quad S(4)=28
- Add the sizes of the three stages.|1-(-4)|+|0-1|+|28-0|=5+1+28=34
\text{total distance}=34
Note. No units are given; the answers are in the units of S and t (for example metres and seconds).
The position of an object moving on a line is given by S=6t^2-t^3, t\ge0, where S is in metres and t is in seconds.
Solution
(i) Determine the velocity and acceleration of the object at t=2.
- Differentiate.v=12t-3t^2,\qquad a=12-6t
- Put t=2.v(2)=24-12=12\,\mathrm{m\,s^{-1}},\qquad a(2)=12-12=0
v(2)=12\,\mathrm{m\,s^{-1}},\quad a(2)=0
(ii) At what time is the object at rest?
- At rest means v=0.12t-3t^2=3t(4-t)=0\ \Rightarrow\ t=0\ \text{or}\ t=4\,\mathrm{s}
t=0\ \text{and}\ t=4\,\mathrm{s}
(iii) In which direction is the object moving at t=5\,s?
- Find the sign of the velocity.v(5)=60-75=-15\,\mathrm{m\,s^{-1}}<0
- A negative velocity means the object moves in the negative direction, that is back towards its starting point.
\text{negative direction}\ (v=-15\,\mathrm{m\,s^{-1}})
(iv) When is the object moving in a positive direction?
- Positive direction means v>0.3t(4-t)>0\ \Rightarrow\ 0<t<4
0<t<4\,\mathrm{s}
(v) When does the object return to its initial position?
- The initial position is S(0)=0. Solve S=0 for t>0.6t^2-t^3=t^2(6-t)=0\ \Rightarrow\ t=6\,\mathrm{s}
t=6\,\mathrm{s}
A particle P moves along the x^+-axis. The acceleration of P in time t seconds, when t\ge0, is a=(3t+5)\,m/s^2 in the positive x-direction. When t=0, the velocity of P is 2\,m/s in the positive x-direction. When t=T, the velocity of P is 6\,m/s. Find the value of T.
Solution
- Integrate the acceleration.v=\int(3t+5)\,dt=\tfrac32t^2+5t+C
- Use v=2 at t=0.C=2,\qquad v=\tfrac32t^2+5t+2
- Put v=6 at t=T and multiply by 2.\tfrac32T^2+5T+2=6\ \Rightarrow\ 3T^2+10T-8=0
- Factorise.(3T-2)(T+4)=0\ \Rightarrow\ T=\tfrac23\ \text{or}\ T=-4
- Time is not negative.T=\tfrac23\,\mathrm{s}
T=\dfrac23\,\mathrm{s}
A particle P moves along the x^+-axis. At time t seconds the velocity of P is v=(3t^2-4t+3)\,m/s. When t=0, P is at the origin O. Find the distance of P from O when P is moving with minimum velocity.
Solution
- Minimum velocity: set \dfrac{dv}{dt}=0.\dfrac{dv}{dt}=6t-4=0\ \Rightarrow\ t=\dfrac23\,\mathrm{s}
- Check it is a minimum.\dfrac{d^2v}{dt^2}=6>0
- Displacement from O: integrate v with x=0 at t=0.x=\int(3t^2-4t+3)\,dt=t^3-2t^2+3t
- Put t=\tfrac23.x=\dfrac{8}{27}-\dfrac{8}{9}+2=\dfrac{8-24+54}{27}=\dfrac{38}{27}\approx1.41\,\mathrm{m}
- The discriminant of v is 16-36<0, so v>0 always: P never turns back and its distance from O equals this displacement.v_{\min}=v\left(\tfrac23\right)=\tfrac43-\tfrac83+3=\tfrac53\,\mathrm{m\,s^{-1}}
\dfrac{38}{27}\,\mathrm{m}\approx1.41\,\mathrm{m}
A particle P moves along the x-axis in a straight line so that, at time t seconds, the velocity of P is v\,m/s, where v=\begin{cases}10t-2t^2, & 0\le t\le6,\\ -\dfrac{432}{t^2}, & t>6.\end{cases} At t=0, P is at the origin O. Find the displacement of P from O when:
Solution
(i) t=6\,s,
- Displacement from O is the definite integral of v from 0.x(6)=\int_0^6(10t-2t^2)\,dt=\left[5t^2-\tfrac23t^3\right]_0^6
- Evaluate.x(6)=180-144=36\,\mathrm{m}
36\,\mathrm{m}
(ii) t=10\,s.
- Add the displacement from 6 s to 10 s, using the second formula.x(10)=36+\int_6^{10}-\dfrac{432}{t^2}\,dt=36+\left[\dfrac{432}{t}\right]_6^{10}
- Evaluate.x(10)=36+(43.2-72)=36-28.8=7.2\,\mathrm{m}
7.2\,\mathrm{m}
A particle P moves along the x^+-axis. At time t seconds the velocity v of P is increasing in the direction of x-axis given by v=\begin{cases}8t-\tfrac32t^2, & 0\le t\le4,\\ 16-2t, & t>4.\end{cases} When t=0, P is at the origin O. Find:
Solution
(i) the greatest speed of P in the interval 0\le t\le4,
- Differentiate and set to zero.\dfrac{dv}{dt}=8-3t=0\ \Rightarrow\ t=\dfrac83
- Compare with the end points of the interval.v\left(\tfrac83\right)=\dfrac{64}{3}-\dfrac32\cdot\dfrac{64}{9}=\dfrac{64}{3}-\dfrac{32}{3}=\dfrac{32}{3},\qquad v(0)=0,\quad v(4)=8
- v\ge0 on the whole interval, so the greatest velocity is also the greatest speed.
\dfrac{32}{3}\approx10.7\,\mathrm{m\,s^{-1}}
(ii) the distance of P from O when t=4,
- v\ge0 on [0,4], so the distance from O is the integral.x(4)=\int_0^4\left(8t-\tfrac32t^2\right)dt=\left[4t^2-\tfrac12t^3\right]_0^4=64-32=32\,\mathrm{m}
32\,\mathrm{m}
(iii) the time at which P is instantaneously at rest for t>4,
- Set the second formula to zero.16-2t=0\ \Rightarrow\ t=8\,\mathrm{s}
t=8\,\mathrm{s}
(iv) the total distance travelled by P in the first 10\,s of its motion.
- From 4 s to 8 s P still moves forwards.\int_4^8(16-2t)\,dt=\left[16t-t^2\right]_4^8=64-48=16\,\mathrm{m}
- From 8 s to 10 s it moves backwards.\int_8^{10}(16-2t)\,dt=\left[16t-t^2\right]_8^{10}=60-64=-4\,\mathrm{m}
- Distance adds the sizes.32+16+4=52\,\mathrm{m}
52\,\mathrm{m}
A particle P moves along the x^+-axis. The acceleration of P at time t seconds is a=(4t-8)\,m/s^2, measured in the increasing direction of x. The velocity of P at time t seconds is v\,m/s. Given that v=6 when t=0, find:
Solution
(i) v in terms of t.
- Integrate and use v(0)=6.v=\int(4t-8)\,dt=2t^2-8t+C,\qquad C=6
- Factorise (useful for part (ii)).v=2t^2-8t+6=2(t-1)(t-3)
v=2t^2-8t+6
(ii) the distance between the two points where P is instantaneously at rest.
- P is at rest where v=0.t=1\,\mathrm{s}\ \text{and}\ t=3\,\mathrm{s}
- Between these times v<0, so P moves in one direction only. The integral gives the displacement.\int_1^3(2t^2-8t+6)\,dt=\left[\tfrac23t^3-4t^2+6t\right]_1^3=0-\tfrac83=-\tfrac83
- The distance between the two points is the size of this displacement.\left|-\tfrac83\right|=\tfrac83\approx2.67\,\mathrm{m}
\dfrac83\,\mathrm{m}
A particle P moves along the x^+-axis and its acceleration after a given instant t is given by a=(4t-9)\,m/s^2, t\ge0. When t=1, P is moving with velocity of -3\,m/s.
Solution
(i) Find the minimum velocity of P.
- Integrate and use v(1)=-3.v=2t^2-9t+C,\qquad 2-9+C=-3\ \Rightarrow\ C=4,\qquad v=2t^2-9t+4
- Minimum velocity where a=\dfrac{dv}{dt}=0 (and \dfrac{da}{dt}=4>0).4t-9=0\ \Rightarrow\ t=\dfrac94
- Evaluate.v\left(\tfrac94\right)=2\left(\tfrac{81}{16}\right)-\tfrac{81}{4}+4=\tfrac{81}{8}-\tfrac{162}{8}+\tfrac{32}{8}=-\tfrac{49}{8}
v_{\min}=-\dfrac{49}{8}=-6.125\,\mathrm{m\,s^{-1}}
(ii) Determine the times when P is instantaneously at rest.
- Solve v=0 by factorising.2t^2-9t+4=(2t-1)(t-4)=0\ \Rightarrow\ t=\tfrac12\,\mathrm{s},\ t=4\,\mathrm{s}
t=\tfrac12\,\mathrm{s}\ \text{and}\ t=4\,\mathrm{s}
(iii) Find the distance travelled by P in the first 4\tfrac12 seconds of its motion.
- Take x(t)=\int v\,dt with x(0)=0.x(t)=\tfrac23t^3-\tfrac92t^2+4t
- Positions at 0, the rest times and 4\tfrac12.x(0)=0,\quad x\left(\tfrac12\right)=\tfrac{23}{24},\quad x(4)=-\tfrac{40}{3},\quad x\left(\tfrac92\right)=-\tfrac{99}{8}
- Add the sizes of the three stages.\tfrac{23}{24}+\left|-\tfrac{40}{3}-\tfrac{23}{24}\right|+\left|-\tfrac{99}{8}+\tfrac{40}{3}\right|=\tfrac{23}{24}+\tfrac{343}{24}+\tfrac{23}{24}=\tfrac{389}{24}
\dfrac{389}{24}\approx16.2\,\mathrm{m}
A car moving on a straight road is modelled as a particle moving along the x^+-axis, and its acceleration a\,m/s^2, after a given instant t, is given by a=\begin{cases}4-\tfrac12t, & 0\le t\le8\\ 0, & t>8\end{cases} The car starts from rest.
Solution
(i) Find a similar expression for the velocity of the car, as that of its acceleration.
- For 0\le t\le8 integrate and use v(0)=0.v=\int\left(4-\tfrac12t\right)dt=4t-\tfrac14t^2
- At t=8 the velocity isv(8)=32-16=16\,\mathrm{m\,s^{-1}}
- For t>8 the acceleration is zero, so the velocity stays at 16.
a-t graph
v-t graph
v=\begin{cases}4t-\tfrac14t^2, & 0\le t\le8\\ 16, & t>8\end{cases}
(ii) Find the time it takes for the car to reach its maximum speed.
- While 0\le t<8 the acceleration 4-\tfrac12t is positive, so the speed keeps increasing. At t=8 the acceleration becomes zero and the speed then stays constant.a=0\ \Rightarrow\ 4-\tfrac12t=0\ \Rightarrow\ t=8\,\mathrm{s}
t=8\,\mathrm{s}\ (\text{maximum speed } 16\,\mathrm{m\,s^{-1}})
(iii) Show that the displacement of the car from O is given by S=\begin{cases}2t^2-\tfrac1{12}t^3, & 0\le t\le8\\ 16t-\tfrac{128}{3}, & t>8\end{cases}
- For 0\le t\le8 integrate v with S(0)=0.S=\int\left(4t-\tfrac14t^2\right)dt=2t^2-\tfrac1{12}t^3
- Position at t=8.S(8)=128-\tfrac{512}{12}=128-\tfrac{128}{3}=\tfrac{256}{3}
- For t>8 the car moves at 16\,\mathrm{m\,s^{-1}} from S(8).S=\tfrac{256}{3}+16(t-8)=16t-128+\tfrac{256}{3}=16t-\tfrac{128}{3}
S=2t^2-\tfrac1{12}t^3\ (0\le t\le8),\qquad S=16t-\tfrac{128}{3}\ (t>8)
(iv) Calculate the time it takes the car to cover the first 1000\,m.
- At t=8 the car has covered only \tfrac{256}{3}\approx85.3\,\mathrm{m}, so use the second formula.16t-\tfrac{128}{3}=1000
- Solve.16t=\tfrac{3128}{3}\ \Rightarrow\ t=\tfrac{3128}{48}=\tfrac{391}{6}\approx65.2\,\mathrm{s}
t=\dfrac{391}{6}\approx65.2\,\mathrm{s}
Note. Misprints corrected: the book prints the second condition as t>0 (it must be t>8) and part (iii) says “displacement of P form O” (read: of the car from O).
A particle is moving with constant acceleration a with an initial velocity v_i and after time t it covers a distance S. Prove that:
Solution
(i) v_f=v_i+at
- Acceleration is the rate of change of velocity, and a is constant.\dfrac{dv}{dt}=a
- Integrate with respect to time.v=\int a\,dt=at+C_1
- At t=0 the velocity is v_i, so C_1=v_i. Call the velocity at time t the final velocity v_f.v_f=v_i+at
v_f=v_i+at
(ii) S=v_it+\tfrac12at^2
- Velocity is the rate of change of displacement; use part (i).\dfrac{dS}{dt}=v_i+at
- Integrate with respect to time (v_i and a are constants).S=v_it+\tfrac12at^2+C_2
- S is measured from the starting point, so S=0 at t=0 and C_2=0.S=v_it+\tfrac12at^2
S=v_it+\tfrac12at^2
(iii) 2aS=v_f^2-v_i^2
- From (i), the time ist=\dfrac{v_f-v_i}{a}
- Substitute in (ii).S=v_i\left(\dfrac{v_f-v_i}{a}\right)+\dfrac12a\left(\dfrac{v_f-v_i}{a}\right)^2
- Multiply both sides by 2a and expand.2aS=2v_iv_f-2v_i^2+v_f^2-2v_iv_f+v_i^2
- The v_iv_f terms cancel.2aS=v_f^2-v_i^2
2aS=v_f^2-v_i^2
Exercise 5.3p. 156
Work component by component: the domain is the intersection of the component domains, derivatives are taken separately for each unit vector, and the magnitude combines the components.
If \mathbf{r}(t) is the position of the particle. Find its domain and the range at given point. Also find its first and second derivative.
Solution
i. \mathbf{r}(t)=(t+1)\hat{\imath}+(t^2-1)\hat{\jmath}+5t\hat{k},\quad t=2
- Domain. All three components are polynomials, defined for every real t.D=\mathbb R
- **Value at t=2.**\mathbf{r}(2)=3\hat{\imath}+3\hat{\jmath}+10\hat{k}
- First derivative: differentiate each component.\mathbf{r}'(t)=\hat{\imath}+2t\hat{\jmath}+5\hat{k},\qquad \mathbf{r}'(2)=\hat{\imath}+4\hat{\jmath}+5\hat{k}
- Second derivative.\mathbf{r}''(t)=2\hat{\jmath},\qquad \mathbf{r}''(2)=2\hat{\jmath}
D=\mathbb R;\ \mathbf{r}(2)=3\hat{\imath}+3\hat{\jmath}+10\hat{k};\ \mathbf{r}'(t)=\hat{\imath}+2t\hat{\jmath}+5\hat{k};\ \mathbf{r}''(t)=2\hat{\jmath}
ii. \mathbf{r}(t)=\dfrac{t}{t+1}\hat{\imath}+\dfrac1t\hat{\jmath}+t^3\hat{k},\quad t=-\dfrac12
- Domain. The denominators t+1 and t must not be zero.D=\mathbb R\setminus\{-1,\,0\}
- **Value at t=-\tfrac12.** Here t+1=\tfrac12.\mathbf{r}\left(-\tfrac12\right)=\dfrac{-1/2}{1/2}\hat{\imath}+(-2)\hat{\jmath}+\left(-\tfrac18\right)\hat{k}=-\hat{\imath}-2\hat{\jmath}-\tfrac18\hat{k}
- First derivative. Quotient rule: \dfrac{d}{dt}\dfrac{t}{t+1}=\dfrac{(t+1)-t}{(t+1)^2}=\dfrac{1}{(t+1)^2}.\mathbf{r}'(t)=\dfrac{1}{(t+1)^2}\hat{\imath}-\dfrac{1}{t^2}\hat{\jmath}+3t^2\hat{k},\qquad \mathbf{r}'\left(-\tfrac12\right)=4\hat{\imath}-4\hat{\jmath}+\tfrac34\hat{k}
- Second derivative.\mathbf{r}''(t)=-\dfrac{2}{(t+1)^3}\hat{\imath}+\dfrac{2}{t^3}\hat{\jmath}+6t\hat{k},\qquad \mathbf{r}''\left(-\tfrac12\right)=-16\hat{\imath}-16\hat{\jmath}-3\hat{k}
D=\mathbb R\setminus\{-1,0\};\ \mathbf{r}\left(-\tfrac12\right)=-\hat{\imath}-2\hat{\jmath}-\tfrac18\hat{k};\ \mathbf{r}'\left(-\tfrac12\right)=4\hat{\imath}-4\hat{\jmath}+\tfrac34\hat{k};\ \mathbf{r}''\left(-\tfrac12\right)=-16\hat{\imath}-16\hat{\jmath}-3\hat{k}
iii. \mathbf{r}(t)=e^t\hat{\imath}+\dfrac29e^{2t}\hat{\jmath}+5e^{-t}\hat{k},\quad t=\ln2
- Domain. Exponential functions are defined for all real t.D=\mathbb R
- **Value at t=\ln2.** Use e^{\ln2}=2, e^{2\ln2}=4, e^{-\ln2}=\tfrac12.\mathbf{r}(\ln2)=2\hat{\imath}+\dfrac89\hat{\jmath}+\dfrac52\hat{k}
- First derivative (chain rule: \dfrac{d}{dt}e^{kt}=ke^{kt}).\mathbf{r}'(t)=e^t\hat{\imath}+\dfrac49e^{2t}\hat{\jmath}-5e^{-t}\hat{k},\qquad \mathbf{r}'(\ln2)=2\hat{\imath}+\dfrac{16}{9}\hat{\jmath}-\dfrac52\hat{k}
- Second derivative.\mathbf{r}''(t)=e^t\hat{\imath}+\dfrac89e^{2t}\hat{\jmath}+5e^{-t}\hat{k},\qquad \mathbf{r}''(\ln2)=2\hat{\imath}+\dfrac{32}{9}\hat{\jmath}+\dfrac52\hat{k}
D=\mathbb R;\ \mathbf{r}(\ln2)=2\hat{\imath}+\tfrac89\hat{\jmath}+\tfrac52\hat{k};\ \mathbf{r}'(\ln2)=2\hat{\imath}+\tfrac{16}{9}\hat{\jmath}-\tfrac52\hat{k};\ \mathbf{r}''(\ln2)=2\hat{\imath}+\tfrac{32}{9}\hat{\jmath}+\tfrac52\hat{k}
iv. \mathbf{r}(t)=(\cos2t)\hat{\imath}+(3\sin2t)\hat{\jmath}+5t\hat{k},\quad t=0
- Domain. Sine, cosine and 5t are defined for all real t.D=\mathbb R
- **Value at t=0.**\mathbf{r}(0)=\cos0\,\hat{\imath}+3\sin0\,\hat{\jmath}+0\hat{k}=\hat{\imath}
- First derivative (chain rule).\mathbf{r}'(t)=-2\sin2t\,\hat{\imath}+6\cos2t\,\hat{\jmath}+5\hat{k},\qquad \mathbf{r}'(0)=6\hat{\jmath}+5\hat{k}
- Second derivative.\mathbf{r}''(t)=-4\cos2t\,\hat{\imath}-12\sin2t\,\hat{\jmath},\qquad \mathbf{r}''(0)=-4\hat{\imath}
D=\mathbb R;\ \mathbf{r}(0)=\hat{\imath};\ \mathbf{r}'(0)=6\hat{\jmath}+5\hat{k};\ \mathbf{r}''(0)=-4\hat{\imath}
Note. “Range at given point” is read as the output \mathbf{r}(t_0) at the given value of t (a single vector). The derivatives are found in general and then evaluated at the same point.
Find the velocity and acceleration of the function at t=0
Solution
i. \mathbf{r}(t)=(3t+1)\hat{\imath}+\sqrt3\,t\hat{\jmath}+t^2\hat{k}
- Velocity is the first derivative.\mathbf{v}(t)=3\hat{\imath}+\sqrt3\hat{\jmath}+2t\hat{k}\ \Rightarrow\ \mathbf{v}(0)=3\hat{\imath}+\sqrt3\hat{\jmath}
- Acceleration is the second derivative.\mathbf{a}(t)=2\hat{k}\ \Rightarrow\ \mathbf{a}(0)=2\hat{k}
\mathbf{v}(0)=3\hat{\imath}+\sqrt3\hat{\jmath},\qquad \mathbf{a}(0)=2\hat{k}
ii. \mathbf{r}(t)=\left(\dfrac{t}{\sqrt2}\right)\hat{\imath}+\left(\dfrac{t}{\sqrt2}-16t^2\right)\hat{\jmath}-2t\hat{k}
- Differentiate each component.\mathbf{v}(t)=\dfrac{1}{\sqrt2}\hat{\imath}+\left(\dfrac1{\sqrt2}-32t\right)\hat{\jmath}-2\hat{k}\ \Rightarrow\ \mathbf{v}(0)=\dfrac1{\sqrt2}\hat{\imath}+\dfrac1{\sqrt2}\hat{\jmath}-2\hat{k}
- Differentiate again; only the \hat{\jmath} component has a t left.\mathbf{a}(t)=-32\hat{\jmath}\ \Rightarrow\ \mathbf{a}(0)=-32\hat{\jmath}
\mathbf{v}(0)=\tfrac1{\sqrt2}\hat{\imath}+\tfrac1{\sqrt2}\hat{\jmath}-2\hat{k},\qquad \mathbf{a}(0)=-32\hat{\jmath}
iii. \mathbf{r}(t)=\left(\ln(t^2+1)\right)\hat{\imath}+\left(\tan^{-1}t\right)\hat{\jmath}+\sqrt{t^2+1}\,\hat{k}
- Differentiate: chain rule for the logarithm and the root, standard result for \tan^{-1}t.\mathbf{v}(t)=\dfrac{2t}{t^2+1}\hat{\imath}+\dfrac{1}{1+t^2}\hat{\jmath}+\dfrac{t}{\sqrt{t^2+1}}\hat{k}\ \Rightarrow\ \mathbf{v}(0)=\hat{\jmath}
- Differentiate again (quotient rule for the first two, product rule for the third).\mathbf{a}(t)=\dfrac{2(1-t^2)}{(t^2+1)^2}\hat{\imath}-\dfrac{2t}{(1+t^2)^2}\hat{\jmath}+\dfrac{1}{(t^2+1)^{3/2}}\hat{k}
- Put t=0.\mathbf{a}(0)=2\hat{\imath}+0\hat{\jmath}+\hat{k}=2\hat{\imath}+\hat{k}
\mathbf{v}(0)=\hat{\jmath},\qquad \mathbf{a}(0)=2\hat{\imath}+\hat{k}
iv. \mathbf{r}(t)=\dfrac49(t+1)^{3/2}\hat{\imath}+\dfrac49(1-t)^{3/2}\hat{\jmath}+\dfrac13t\hat{k}
- Power and chain rules: \dfrac{d}{dt}(1-t)^{3/2}=-\tfrac32(1-t)^{1/2}.\mathbf{v}(t)=\dfrac23(t+1)^{1/2}\hat{\imath}-\dfrac23(1-t)^{1/2}\hat{\jmath}+\dfrac13\hat{k}\ \Rightarrow\ \mathbf{v}(0)=\dfrac23\hat{\imath}-\dfrac23\hat{\jmath}+\dfrac13\hat{k}
- Differentiate again.\mathbf{a}(t)=\dfrac13(t+1)^{-1/2}\hat{\imath}+\dfrac13(1-t)^{-1/2}\hat{\jmath}\ \Rightarrow\ \mathbf{a}(0)=\dfrac13\hat{\imath}+\dfrac13\hat{\jmath}
- Note: |\mathbf{v}(0)|=\sqrt{\tfrac49+\tfrac49+\tfrac19}=1, so the particle moves with unit speed at t=0.
\mathbf{v}(0)=\tfrac23\hat{\imath}-\tfrac23\hat{\jmath}+\tfrac13\hat{k},\qquad \mathbf{a}(0)=\tfrac13\hat{\imath}+\tfrac13\hat{\jmath}
Find the scalar valued function in term of the magnitude
Solution
i. \mathbf{r}(t)=(3t-7)\hat{\imath}+t\hat{\jmath}-t^2\hat{k}
- Magnitude: square root of the sum of the squares of the components.|\mathbf{r}(t)|=\sqrt{(3t-7)^2+t^2+(-t^2)^2}
- Expand and collect.|\mathbf{r}(t)|=\sqrt{9t^2-42t+49+t^2+t^4}=\sqrt{t^4+10t^2-42t+49}
|\mathbf{r}(t)|=\sqrt{t^4+10t^2-42t+49}
ii. \mathbf{r}(t)=\left(\dfrac{t}{\sqrt2}\right)\hat{\imath}+\left(\dfrac{t}{\sqrt2}+16t^2\right)\hat{\jmath}-2t\hat{k}
- Square each component.\left(\dfrac{t}{\sqrt2}\right)^2=\dfrac{t^2}{2},\quad \left(\dfrac{t}{\sqrt2}+16t^2\right)^2=\dfrac{t^2}{2}+16\sqrt2\,t^3+256t^4,\quad (-2t)^2=4t^2
- Add and take the square root.|\mathbf{r}(t)|=\sqrt{256t^4+16\sqrt2\,t^3+5t^2}=|t|\sqrt{256t^2+16\sqrt2\,t+5}
|\mathbf{r}(t)|=\sqrt{256t^4+16\sqrt2\,t^3+5t^2}
iii. \mathbf{r}(t)=\left(\ln(t^2+1)\right)\hat{\imath}+(\tan t)\hat{\jmath}+\sec t\,\hat{k}
- Square and add the components.|\mathbf{r}(t)|=\sqrt{\left[\ln(t^2+1)\right]^2+\tan^2t+\sec^2t}
- Use \sec^2t=1+\tan^2t to simplify.|\mathbf{r}(t)|=\sqrt{\left[\ln(t^2+1)\right]^2+2\tan^2t+1}
- It is defined wherever \tan t and \sec t are, that is t\ne\dfrac{\pi}{2}+n\pi.
|\mathbf{r}(t)|=\sqrt{\left[\ln(t^2+1)\right]^2+2\tan^2t+1}
iv. \mathbf{r}(t)=\dfrac49(t+1)^{3/2}\hat{\imath}+\dfrac49(1-t)^{3/2}\hat{\jmath}+\dfrac13t^{3/2}\hat{k}
- Square each component.|\mathbf{r}|^2=\dfrac{16}{81}(t+1)^3+\dfrac{16}{81}(1-t)^3+\dfrac19t^3
- Simplify with (1+t)^3+(1-t)^3=2+6t^2.|\mathbf{r}|^2=\dfrac{16}{81}(2+6t^2)+\dfrac{9t^3}{81}=\dfrac{9t^3+96t^2+32}{81}
- Take the square root. The components need t+1\ge0, 1-t\ge0 and t\ge0, so 0\le t\le1.|\mathbf{r}(t)|=\dfrac19\sqrt{9t^3+96t^2+32},\qquad 0\le t\le1
|\mathbf{r}(t)|=\dfrac19\sqrt{9t^3+96t^2+32}
Review exercisep. 156–159
Mixed revision: scalars and vectors, reading motion graphs, and describing motion from v(t).
Which of the following quantities is a vector?
- ACharge
- BMass
- CMomentumAnswer
- DTime
Why
Momentum is mass × velocity. Velocity has a direction, so momentum has the same direction and is a vector. Charge, mass and time have magnitude only.
Which of following is scalar quantity?
- ADisplacement
- BWeight
- CForce
- DWorkAnswer
Why
Work is the dot product of force and displacement, W=\mathbf F\cdot\mathbf d, and a dot product is a number. Displacement, weight and force all have direction.
Which of following can be used to determine the magnitude of velocity?
- AArea under acceleration-time graphAnswer
- BArea under velocity-time graph
- CGradient of an acceleration-time graph
- DGradient of a velocity-time graph
Why
Since v=\int a\,dt, the area under an a-t graph gives the change in velocity, and so the velocity when the starting velocity is known (for example from rest).
The area under a v-t graph gives displacement, and the gradient of a v-t graph gives acceleration, so (B) and (D) are tempting but give other quantities.
The winner of 400 metre race must have the greatest:
- Aacceleration
- Baverage speedAnswer
- Cinstantaneous speed
- Dmaximum speed
Why
Every runner covers the same 400 m, and the winner takes the least time, so the winner has the greatest average speed =\dfrac{400}{\text{time}}. Another runner may reach a higher top speed or acceleration for a moment and still lose.
A car travels 100\,km. The journey takes two hours. The highest speed of the car is 80\,km/h, and the lowest speed 40\,km/h. What is the average speed for the journey?
- A40\,km/h
- B50\,km/hAnswer
- C60\,km/h
- D120\,km/h
Why
Average speed = total distance ÷ total time =\dfrac{100}{2}=50\,\mathrm{km/h}. The highest and lowest speeds are not needed; (C) is their mean, which is a common wrong method.
A car travels 6.0\,km along a main road in 6.0 minutes. It then travels 2\,km along minor road in 6.0 minutes. Which calculation of average speed for the whole journey is correct?
- A\dfrac{8.0}{12.0}=0.67\,\dfrac{km}{min}Answer
- B\dfrac{12.0}{8.0}=1.5\,\dfrac{km}{min}
- C8.0+12.0=20\,\dfrac{km}{min}
- D8.0\times12.0=96\,\dfrac{km}{min}
Why
Total distance =6.0+2.0=8.0\,\mathrm{km} and total time =6.0+6.0=12.0\,\mathrm{min}, so the average speed is \dfrac{8.0}{12.0}=0.67\,\mathrm{km/min}. Option (B) divides time by distance, the wrong way round.
Which person is experiencing an acceleration?
- AA driver of a car that is braking to stop at traffic light.Answer
- BA passenger in a train that is stationary in a railway station.
- CA shopper in a large store ascending an escalator (moving stairs) at a uniform speed.
- DA skydiver that is falling at a constant speed towards the Earth.
Why
Acceleration means the velocity is changing. The braking car's speed is decreasing, so its driver has a (negative) acceleration. In (B) the velocity is zero, and in (C) and (D) the velocity is constant, so the acceleration is zero.
A child is standing on the platform of station. A train travelling at 30\,m/s takes 3.0\,s to pass the child. What is length of train?
- A10\,m
- B27\,m
- C30\,m
- D90\,mAnswer
Why
While passing the child, the train moves forward by its own length: length = speed × time =30\times3.0=90\,\mathrm{m}.
A heavy object is released near the surface of earth and falls freely. Air resistance can be ignored. Which statement about the acceleration of the object due to gravity is correct?
- AThe acceleration depends on mass of the object.
- BThe acceleration depends on volume of the object.
- CThe acceleration is constant.Answer
- DThe acceleration is initially zero and increases as the object falls.
Why
Without air resistance every body falls with the same constant acceleration g=9.8\,\mathrm{m\,s^{-2}}, whatever its mass or volume. It is g from the moment of release, not zero.
The graph shows the movement of a car from point A to point D, D is 500\,m from A, note that there are two slant lines AB and CD. The slant lines indicate that car is moving. The flat line BC indicates that car is stopped or is at rest.
Solution
(i) Calculate the speed of car during the first 10 seconds.
- Slope of AB: the car covers 200\,\mathrm{m} in 10\,\mathrm{s}.v=\dfrac{200-0}{10-0}=20\,\mathrm{m\,s^{-1}}
20\,\mathrm{m\,s^{-1}}
(ii) For how long did the car stop?
- The line BC is flat from t=10\,\mathrm{s} to t=30\,\mathrm{s}.30-10=20\,\mathrm{s}
20\,\mathrm{s}
(iii) What is speed of car on its journey from C to D?
- Slope of CD: from 200\,\mathrm{m} to 500\,\mathrm{m} between 30\,\mathrm{s} and 60\,\mathrm{s}.v=\dfrac{500-200}{60-30}=\dfrac{300}{30}=10\,\mathrm{m\,s^{-1}}
10\,\mathrm{m\,s^{-1}}
(iv) On which part of the journey did the car travel faster?
- Compare the slopes: AB has slope 20 and CD has slope 10. The steeper line means the greater speed.
\text{on AB }(20\,\mathrm{m\,s^{-1}})
(v) What is average speed of the car for whole journey?
- Total distance 500\,\mathrm{m} in total time 60\,\mathrm{s}, including the stop.\dfrac{500}{60}\approx8.33\,\mathrm{m\,s^{-1}}
8.33\,\mathrm{m\,s^{-1}}
(vi) What is the average speed of the car for the time it was moving?
- Moving time =10+30=40\,\mathrm{s} (leave out the 20 s stop).\dfrac{500}{40}=12.5\,\mathrm{m\,s^{-1}}
12.5\,\mathrm{m\,s^{-1}}
The graph below shows how the speed of an athlete varies during a race.
Solution
(i) Calculate the acceleration of athlete during the first 4 second.
- Slope of the first line.a=\dfrac{8-0}{4-0}=2\,\mathrm{m\,s^{-2}}
2\,\mathrm{m\,s^{-2}}
(ii) What was the athlete doing between the 4^{th} and 14^{th} second?
The graph is horizontal from t=4\,\mathrm{s} to t=14\,\mathrm{s}, so the athlete ran at a constant speed of 8\,\mathrm{m\,s^{-1}} with zero acceleration. In these 10 seconds he covered 8\times10=80\,\mathrm{m}.
\text{running at a steady } 8\,\mathrm{m\,s^{-1}}
(iii) Calculate the deceleration of the athlete in final stage of the race.
- Slope of the last line is negative; its size is the deceleration.\text{slope}=\dfrac{0-8}{16-14}=-4\ \Rightarrow\ \text{deceleration}=4\,\mathrm{m\,s^{-2}}
4\,\mathrm{m\,s^{-2}}
(iv) How far has the athlete moved in the first 4 seconds?
- Area of the triangle under the first line.\tfrac12(4)(8)=16\,\mathrm{m}
16\,\mathrm{m}
(v) What is the total distance travelled by the athlete?
- Area of the whole trapezium: parallel sides 10\,\mathrm{s} (top) and 16\,\mathrm{s} (bottom), height 8.\tfrac12(10+16)(8)=104\,\mathrm{m}
- Check by pieces.16+80+\tfrac12(2)(8)=16+80+8=104\,\mathrm{m}
104\,\mathrm{m}
Let v(t)=\dfrac1\pi+\sin3t represent the velocity of an object moving on a line. At t=\dfrac{\pi}{3} the position is 4.
Solution
(i) Write the acceleration function.
- Differentiate the velocity (chain rule).a(t)=\dfrac{d}{dt}\left(\dfrac1\pi+\sin3t\right)=3\cos3t
a(t)=3\cos3t
(ii) Write the position function.
- Integrate the velocity.s(t)=\int\left(\dfrac1\pi+\sin3t\right)dt=\dfrac{t}{\pi}-\dfrac13\cos3t+C
- Use s\left(\tfrac\pi3\right)=4; note \cos\pi=-1.4=\dfrac13-\dfrac13\cos\pi+C=\dfrac13+\dfrac13+C\ \Rightarrow\ C=\dfrac{10}{3}
s(t)=\dfrac{t}{\pi}-\dfrac13\cos3t+\dfrac{10}{3}
(iii) At t=\dfrac{\pi}{4}, is the object speeding up or slowing down? Explain your answer.
- Find the signs of v and a at t=\tfrac\pi4; \sin\tfrac{3\pi}{4}=\tfrac{\sqrt2}{2} and \cos\tfrac{3\pi}{4}=-\tfrac{\sqrt2}{2}.v\left(\tfrac\pi4\right)=\dfrac1\pi+\dfrac{\sqrt2}{2}\approx1.03>0,\qquad a\left(\tfrac\pi4\right)=-\dfrac{3\sqrt2}{2}\approx-2.12<0
- Velocity and acceleration have opposite signs, so the speed is decreasing.
\text{slowing down, since } v>0 \text{ and } a<0
(iv) On the interval \left[\dfrac{\pi}{2},\pi\right], what is the velocity of object when acceleration is 3?
- Set a=3.3\cos3t=3\ \Rightarrow\ \cos3t=1\ \Rightarrow\ 3t=2n\pi\ \Rightarrow\ t=\dfrac{2n\pi}{3}
- The only such t in \left[\tfrac\pi2,\pi\right] is n=1.t=\dfrac{2\pi}{3}
- Velocity at that time; \sin2\pi=0.v\left(\tfrac{2\pi}{3}\right)=\dfrac1\pi+\sin2\pi=\dfrac1\pi\approx0.318
v=\dfrac1\pi\approx0.318
A particle moves along y-axis so that its velocity v at time t\ge0 is given by v(t)=1-\tan^{-1}(e^t). At time t=0 the particle is at y=-1.
Solution
(a) Find acceleration of particle at time t=2.
- Differentiate: \dfrac{d}{dt}\tan^{-1}u=\dfrac{1}{1+u^2}\dfrac{du}{dt} with u=e^t.a(t)=-\dfrac{e^t}{1+e^{2t}}
- Put t=2.a(2)=-\dfrac{e^2}{1+e^4}=-\dfrac{7.389}{55.598}\approx-0.133
a(2)=-\dfrac{e^2}{1+e^4}\approx-0.133
(b) Is the speed of particle increasing or decreasing at time t=2? Give a reason for your answer.
- Velocity at t=2 (radians).v(2)=1-\tan^{-1}(e^2)=1-\tan^{-1}(7.389)=1-1.436\approx-0.436<0
- v(2)<0 and a(2)<0 have the same sign, so the speed is increasing.
\text{increasing, since } v(2) \text{ and } a(2) \text{ are both negative}
(c) Find the time t\ge0 at which the particle reaches its highest point. Justify your answer.
- The highest point can only occur where v=0 (or at t=0).1-\tan^{-1}(e^t)=0\ \Rightarrow\ \tan^{-1}(e^t)=1\ \Rightarrow\ e^t=\tan1\ \Rightarrow\ t=\ln(\tan1)\approx0.443
- \tan^{-1}(e^t) increases with t, so v>0 for 0\le t<\ln(\tan1) and v<0 after it. The particle moves up and then down for ever, so y is greatest at t=\ln(\tan1).
t=\ln(\tan1)\approx0.443
Note. Use radians. The starting position y=-1 is not needed for these three parts.
FBISE 2026 questions
Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.
For the vector function \mathbf{r}(t)=\langle\ln(t-1),\sqrt{4-t}\rangle, the domain in \mathbb R is:
- A(1,4)
- B[1,4)
- C(1,4]Answer
- D[1,4]
Why
\ln(t-1) needs t-1>0, so t>1 (1 is excluded). \sqrt{4-t} needs 4-t\ge0, so t\le4 (4 is included). The domain is the intersection (1,4].
(B) and (D) wrongly include t=1, where \ln0 is not defined; (A) wrongly leaves out t=4, where \sqrt0=0 is fine.
The position of a drone is \mathbf{r}(t)=\langle5t,\,4t-t^2,\,2\rangle (meters, t in seconds). The acceleration at t=2 seconds is:
- A\langle0,-2,0\rangleAnswer
- B\langle0,2,0\rangle
- C\langle5,-2,0\rangle
- D\langle5,2,0\rangle
Why
\mathbf{v}(t)=\langle5,\,4-2t,\,0\rangle and \mathbf{a}(t)=\langle0,\,-2,\,0\rangle, the same at every time, including t=2.
(C) keeps the 5 from the first derivative; the constant 5 differentiates to 0.
The displacement–time graph of a particle is a straight line from (0,2) to (5,12), where displacement is in meters and time in seconds.
Solution
(a) Find the velocity of the particle.
- The velocity is the slope of the s-t graph, which is constant for a straight line.v=\dfrac{s_2-s_1}{t_2-t_1}=\dfrac{12-2}{5-0}=\dfrac{10}{5}=2\,\mathrm{m\,s^{-1}}
The line is straight, so the particle moves with this velocity at every instant (uniform motion).
v=2\,\mathrm{m\,s^{-1}}
(b) Sketch the displacement–time graph.
Plot (0,\,2) and (5,\,12) with time on the horizontal axis and displacement on the vertical axis, and join them with a straight line.
Displacement-time graph
(c) Sketch the corresponding velocity–time graph.
The velocity is the same, 2\,\mathrm{m\,s^{-1}}, from t=0 to t=5, so the v-t graph is a horizontal line segment. Check: the area under it, 2\times5=10\,\mathrm{m}, equals the change in displacement 12-2=10\,\mathrm{m}.
Velocity-time graph
Given \mathbf{r}(t)=\langle t^2-1,\,3t,\,\ln(t+1)\rangle, find \mathbf{r}'(t) and the magnitude of \mathbf{r}'(2).
Solution
- Differentiate each component.\mathbf{r}'(t)=\left\langle 2t,\ 3,\ \dfrac{1}{t+1}\right\rangle
- Put t=2.\mathbf{r}'(2)=\left\langle4,\ 3,\ \tfrac13\right\rangle
- Magnitude: square root of the sum of squares.|\mathbf{r}'(2)|=\sqrt{16+9+\tfrac19}=\sqrt{\dfrac{226}{9}}=\dfrac{\sqrt{226}}{3}\approx5.01
\mathbf{r}'(t)=\left\langle2t,3,\tfrac{1}{t+1}\right\rangle,\qquad |\mathbf{r}'(2)|=\dfrac{\sqrt{226}}{3}\approx5.01
The velocity of a particle is v(t)=3t^2-12t+9\,\mathrm{m/s}.
Solution
(a) Find displacement from t=0 to t=5 seconds.
- Displacement is the definite integral of velocity.\int_0^5(3t^2-12t+9)\,dt=\left[t^3-6t^2+9t\right]_0^5
- Evaluate.=125-150+45=20\,\mathrm{m}
20\,\mathrm{m}
(b) Find total distance travelled in the same interval.
- Find where the particle stops: factorise v.3t^2-12t+9=3(t-1)(t-3)=0\ \Rightarrow\ t=1,\ t=3
- With s(t)=t^3-6t^2+9t (taking s(0)=0), find the positions.s(0)=0,\quad s(1)=4,\quad s(3)=0,\quad s(5)=20
- Add the sizes of the three stages.|4-0|+|0-4|+|20-0|=4+4+20=28\,\mathrm{m}
28\,\mathrm{m}
A car moves along a curve with position \mathbf{r}(t)=\langle t^2-2t,\,t^3-3t\rangle.
Solution
(a) Find the velocity vector at t=1.
- Velocity is the derivative of position.\mathbf{v}(t)=\mathbf{r}'(t)=\langle2t-2,\ 3t^2-3\rangle
- Put t=1.\mathbf{v}(1)=\langle2-2,\ 3-3\rangle=\langle0,\ 0\rangle
\mathbf{v}(1)=\langle0,0\rangle=\mathbf 0
(b) Find the speed at t=1.
- Speed is the magnitude of the velocity.|\mathbf{v}(1)|=\sqrt{0^2+0^2}=0
The car is momentarily at rest at t=1, at the point \mathbf{r}(1)=\langle-1,-2\rangle. The path has a sharp point (cusp) there: the car stops and sets off again along the same tangent line.
\text{speed}=0
A particle moves such that its position vector is \mathbf{r}(t)=4\cos(2t)\,\hat{\imath}+4\sin(2t)\,\hat{\jmath}
Solution
(a) Find the velocity and acceleration vectors.
- Velocity: differentiate each component (chain rule, the inner derivative is 2).\mathbf{v}(t)=\mathbf{r}'(t)=-8\sin(2t)\,\hat{\imath}+8\cos(2t)\,\hat{\jmath}
- Acceleration: differentiate again.\mathbf{a}(t)=\mathbf{v}'(t)=-16\cos(2t)\,\hat{\imath}-16\sin(2t)\,\hat{\jmath}
\mathbf{v}=-8\sin2t\,\hat{\imath}+8\cos2t\,\hat{\jmath},\quad \mathbf{a}=-16\cos2t\,\hat{\imath}-16\sin2t\,\hat{\jmath}
(b) Show that the acceleration is directed opposite to the position vector.
- Take out the common factor -4 from \mathbf{a}.\mathbf{a}(t)=-4\left[4\cos(2t)\,\hat{\imath}+4\sin(2t)\,\hat{\jmath}\right]
- The bracket is exactly \mathbf{r}(t).\mathbf{a}(t)=-4\,\mathbf{r}(t)
- A vector multiplied by a negative number points the opposite way. So \mathbf{a} is parallel to \mathbf{r} but opposite in direction: it always points from the particle towards the centre O. (Check: \mathbf{a}\cdot\mathbf{r}=-4|\mathbf{r}|^2<0 and |\mathbf{a}\cdot\mathbf{r}|=|\mathbf{a}||\mathbf{r}|, so the angle between them is 180^\circ.)
\mathbf{a}=-4\,\mathbf{r}
(c) Find the speed of the particle and the magnitude of its acceleration.
- Speed is |\mathbf{v}|; use \sin^2+\cos^2=1.|\mathbf{v}|=\sqrt{8^2\sin^2(2t)+8^2\cos^2(2t)}=8
- Magnitude of the acceleration.|\mathbf{a}|=\sqrt{16^2\cos^2(2t)+16^2\sin^2(2t)}=16
- Both are constant. Check with R=4, \omega=2: speed R\omega=8 and |\mathbf{a}|=R\omega^2=\dfrac{v^2}{R}=16.
\text{speed}=8,\qquad |\mathbf{a}|=16
For the vector function \mathbf{r}(t)=\langle\sqrt{t+3},\,\ln(6-t)\rangle, the domain in \mathbb R is:
- A[-3,6]
- B(-3,6)
- C[-3,6)Answer
- D(-3,6]
Why
\sqrt{t+3} needs t\ge-3 (included). \ln(6-t) needs 6-t>0, so t<6 (excluded). The intersection is [-3,6).
The position of a particle is \mathbf{r}(t)=\langle8t,\,6t-2t^2,\,5\rangle (meters, t in seconds). The acceleration at t=3 seconds is:
- A\langle0,-4,0\rangleAnswer
- B\langle8,-4,0\rangle
- C\langle8,0,0\rangle
- D\langle0,4,0\rangle
Why
\mathbf{v}(t)=\langle8,\,6-4t,\,0\rangle and \mathbf{a}(t)=\langle0,\,-4,\,0\rangle, constant for all t, so at t=3 it is \langle0,-4,0\rangle.
The displacement–time graph of a particle is a straight line from (1,-4) to (6,-11), where displacement is in meters and time in seconds.
Solution
(a) Find the velocity of the particle.
- The velocity is the slope of the s-t graph, which is constant for a straight line.v=\dfrac{s_2-s_1}{t_2-t_1}=\dfrac{-11-(-4)}{6-1}=\dfrac{-7}{5}=-1.4\,\mathrm{m\,s^{-1}}
The line is straight, so the particle moves with this velocity at every instant (uniform motion). The negative sign means it moves in the negative direction.
v=-1.4\,\mathrm{m\,s^{-1}}
(b) Sketch the displacement–time graph.
Plot (1,\,-4) and (6,\,-11) with time on the horizontal axis and displacement on the vertical axis, and join them with a straight line.
Displacement-time graph
(c) Sketch the corresponding velocity–time graph.
The velocity is the same, -1.4\,\mathrm{m\,s^{-1}}, from t=1 to t=6, so the v-t graph is a horizontal line segment. Check: the signed area, -1.4\times5=-7\,\mathrm{m} (below the axis), equals the change in displacement -11-(-4)=-7\,\mathrm{m}.
Velocity-time graph
Given the vector function \mathbf{r}(t)=\langle\sqrt{t+4},\,\tan^{-1}(t),\,2^t\rangle find \mathbf{r}'(t) and the magnitude of \mathbf{r}'(3).
Solution
- Differentiate each component: \dfrac{d}{dt}\sqrt{u}=\dfrac{u'}{2\sqrt u}, \dfrac{d}{dt}\tan^{-1}t=\dfrac1{1+t^2}, \dfrac{d}{dt}2^t=2^t\ln2.\mathbf{r}'(t)=\left\langle\dfrac{1}{2\sqrt{t+4}},\ \dfrac{1}{1+t^2},\ 2^t\ln2\right\rangle
- Put t=3.\mathbf{r}'(3)=\left\langle\dfrac{1}{2\sqrt7},\ \dfrac{1}{10},\ 8\ln2\right\rangle\approx\langle0.189,\ 0.1,\ 5.545\rangle
- Magnitude.|\mathbf{r}'(3)|=\sqrt{\dfrac1{28}+\dfrac1{100}+64(\ln2)^2}=\sqrt{0.0357+0.01+30.749}\approx\sqrt{30.795}\approx5.55
\mathbf{r}'(t)=\left\langle\tfrac{1}{2\sqrt{t+4}},\tfrac{1}{1+t^2},2^t\ln2\right\rangle,\qquad |\mathbf{r}'(3)|\approx5.55
The velocity of a particle is v(t)=4t^2-16t+12\,\mathrm{m/s}.
Solution
(a) Find displacement from t=0 to t=3 seconds.
- Integrate the velocity.\int_0^3(4t^2-16t+12)\,dt=\left[\tfrac43t^3-8t^2+12t\right]_0^3
- Evaluate.=36-72+36=0\,\mathrm{m}
- The particle is back at its starting point at t=3.
0\,\mathrm{m}
(b) Find total distance travelled in the same interval.
- Find where v=0.4t^2-16t+12=4(t-1)(t-3)=0\ \Rightarrow\ t=1,\ t=3
- The particle turns back at t=1. With s(t)=\tfrac43t^3-8t^2+12t:s(0)=0,\quad s(1)=\tfrac43-8+12=\tfrac{16}{3},\quad s(3)=0
- Add the sizes of the two stages.\tfrac{16}{3}+\tfrac{16}{3}=\tfrac{32}{3}\approx10.67\,\mathrm{m}
\dfrac{32}{3}\approx10.67\,\mathrm{m}
A particle moves along a curve with position vector: \mathbf{r}(t)=\langle t^3-4t,\,t^2+t\rangle
Solution
(a) Find the velocity vector at t=1.
- Differentiate each component.\mathbf{v}(t)=\langle3t^2-4,\ 2t+1\rangle
- Put t=1.\mathbf{v}(1)=\langle-1,\ 3\rangle=-\hat{\imath}+3\hat{\jmath}
\mathbf{v}(1)=\langle-1,3\rangle
(b) Find the speed at t=1.
- Magnitude of the velocity.|\mathbf{v}(1)|=\sqrt{(-1)^2+3^2}=\sqrt{10}\approx3.16
\sqrt{10}\approx3.16\ \text{units per second}
A particle moves such that its position vector is \mathbf{r}(t)=2\cos(6t)\,\hat{\imath}+2\sin(6t)\,\hat{\jmath}
Solution
(a) Find the velocity and acceleration vectors.
- Velocity: differentiate each component (chain rule, the inner derivative is 6).\mathbf{v}(t)=\mathbf{r}'(t)=-12\sin(6t)\,\hat{\imath}+12\cos(6t)\,\hat{\jmath}
- Acceleration: differentiate again.\mathbf{a}(t)=\mathbf{v}'(t)=-72\cos(6t)\,\hat{\imath}-72\sin(6t)\,\hat{\jmath}
\mathbf{v}=-12\sin6t\,\hat{\imath}+12\cos6t\,\hat{\jmath},\quad \mathbf{a}=-72\cos6t\,\hat{\imath}-72\sin6t\,\hat{\jmath}
(b) Show that the acceleration is directed opposite to the position vector.
- Take out the common factor -36 from \mathbf{a}.\mathbf{a}(t)=-36\left[2\cos(6t)\,\hat{\imath}+2\sin(6t)\,\hat{\jmath}\right]
- The bracket is exactly \mathbf{r}(t).\mathbf{a}(t)=-36\,\mathbf{r}(t)
- A vector multiplied by a negative number points the opposite way. So \mathbf{a} is parallel to \mathbf{r} but opposite in direction: it always points from the particle towards the centre O. (Check: \mathbf{a}\cdot\mathbf{r}=-36|\mathbf{r}|^2<0 and |\mathbf{a}\cdot\mathbf{r}|=|\mathbf{a}||\mathbf{r}|, so the angle between them is 180^\circ.)
\mathbf{a}=-36\,\mathbf{r}
(c) Find the speed of the particle and the magnitude of its acceleration.
- Speed is |\mathbf{v}|; use \sin^2+\cos^2=1.|\mathbf{v}|=\sqrt{12^2\sin^2(6t)+12^2\cos^2(6t)}=12
- Magnitude of the acceleration.|\mathbf{a}|=\sqrt{72^2\cos^2(6t)+72^2\sin^2(6t)}=72
- Both are constant. Check with R=2, \omega=6: speed R\omega=12 and |\mathbf{a}|=R\omega^2=\dfrac{v^2}{R}=72.
\text{speed}=12,\qquad |\mathbf{a}|=72
If \vec f(t)=e^t\hat{\imath}+t^3\hat{\jmath}, then \left[\dfrac{df}{dt}\right]_{t=0} is:
- A\hat{\imath}Answer
- B\hat{\jmath}
- C\hat{\imath}+\hat{\jmath}
- D\hat{\imath}+3\hat{\jmath}
Why
\vec f\,'(t)=e^t\hat{\imath}+3t^2\hat{\jmath}. At t=0: e^0\hat{\imath}+0\hat{\jmath}=\hat{\imath}.
If position of a particle is \vec r(t)=(t^2+1)\hat{\imath}+4t\hat{\jmath}, then magnitude of its acceleration at time t=2 is:
- A0
- B1
- C2Answer
- D4
Why
\mathbf{v}=2t\hat{\imath}+4\hat{\jmath} and \mathbf{a}=2\hat{\imath}, so |\mathbf{a}|=2 at every time, including t=2.
Evaluate \displaystyle\lim_{t\to1}[f(t)\times g(t)] with f(t)=(t^2-1)\hat{\imath}+\sqrt t\,\hat{\jmath}+5\hat{k} and g(t)=(\sin\pi t)\hat{\imath}+e^{t-1}\hat{\jmath}+3t\hat{k}.
Solution
- All components are continuous at t=1, so the limit of the cross product is the cross product of the limits. First find the limits.\lim_{t\to1}f(t)=0\hat{\imath}+1\hat{\jmath}+5\hat{k},\qquad \lim_{t\to1}g(t)=(\sin\pi)\hat{\imath}+e^0\hat{\jmath}+3\hat{k}=0\hat{\imath}+1\hat{\jmath}+3\hat{k}
- Cross product by the determinant.\begin{vmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\0&1&5\\0&1&3\end{vmatrix}=\hat{\imath}(3-5)-\hat{\jmath}(0-0)+\hat{k}(0-0)
- Simplify.\lim_{t\to1}[f(t)\times g(t)]=-2\hat{\imath}
-2\hat{\imath}
Find domain of each component and overall domain of the vector function f(t)=(t+1)\hat{\imath}+\dfrac{1}{t-2}\hat{\jmath}+\sqrt{t+3}\,\hat{k}
Solution
- x(t)=t+1 is a polynomial.D_x=\mathbb R=(-\infty,\infty)
- y(t)=\dfrac1{t-2} needs t\ne2.D_y=\mathbb R\setminus\{2\}=(-\infty,2)\cup(2,\infty)
- z(t)=\sqrt{t+3} needs t+3\ge0.D_z=[-3,\infty)
- The overall domain is the intersection.D_f=D_x\cap D_y\cap D_z=[-3,2)\cup(2,\infty)
D_f=[-3,2)\cup(2,\infty)
A particle is moving in 3D space with position vector r(t)=(2t^2+1)\hat{\imath}+3e^{2t}\hat{\jmath}+(4\cos3t)\hat{k} where t is time in seconds. Find velocity, speed, direction and acceleration of the particle at time t=0 sec.
Solution
Velocity
- Differentiate each component (chain rule for e^{2t} and \cos3t).\mathbf{v}(t)=\mathbf{r}'(t)=4t\hat{\imath}+6e^{2t}\hat{\jmath}-12\sin3t\,\hat{k}
- Put t=0: e^0=1, \sin0=0.\mathbf{v}(0)=0\hat{\imath}+6\hat{\jmath}+0\hat{k}=6\hat{\jmath}
Speed
- Magnitude of the velocity.|\mathbf{v}(0)|=\sqrt{0^2+6^2+0^2}=6\ \text{units per second}
Direction
- Unit vector along the velocity.\dfrac{\mathbf{v}(0)}{|\mathbf{v}(0)|}=\dfrac{6\hat{\jmath}}{6}=\hat{\jmath}
- At t=0 the particle moves in the direction of the positive y-axis.
Acceleration
- Differentiate the velocity.\mathbf{a}(t)=4\hat{\imath}+12e^{2t}\hat{\jmath}-36\cos3t\,\hat{k}
- Put t=0: \cos0=1.\mathbf{a}(0)=4\hat{\imath}+12\hat{\jmath}-36\hat{k},\qquad |\mathbf{a}(0)|=\sqrt{16+144+1296}=4\sqrt{91}\approx38.2
\mathbf{v}(0)=6\hat{\jmath},\quad \text{speed}=6,\quad \text{direction } \hat{\jmath},\quad \mathbf{a}(0)=4\hat{\imath}+12\hat{\jmath}-36\hat{k}
Exam focus
- Uniform circular motion \mathbf{r}=R\cos\omega t\,\hat{\imath}+R\sin\omega t\,\hat{\jmath}: find \mathbf{v} and \mathbf{a}, show \mathbf{a}=-\omega^2\mathbf{r}, find the speed and |\mathbf{a}|. (long)The 8-mark Section C question (Q.6 OR) in both 2026 new-syllabus papers.Go to answer
- Displacement and total distance from a velocity function v(t) over an interval. (problem)Q.2 (ix) in both 2026 sets; the same method is Exercise 5.2 Q.2, Q.9 and Q.11.Go to answer
- Velocity vector and speed of \mathbf{r}(t) at a given time; direction as a unit vector. (problem)Q.2 (xii) in both 2026 sets and the 8-mark old-syllabus Q.4.Go to answer
- Find \mathbf{r}'(t) and the magnitude of \mathbf{r}'(t_0). (problem)Q.2 (vi) OR in both 2026 sets.Go to answer
- Domain of a vector valued function (closed end for a square root, open end for a logarithm). (mcq)MCQ in both 2026 sets and old-syllabus Q.2 (vii); book Example 6.Go to answer
- Straight-line s-t graph: find the velocity and sketch the s-t and v-t graphs. (problem)Q.2 (ii) OR in both 2026 sets.Go to answer
- Acceleration vector from a position vector. (mcq)MCQ 17 and 18 of the 2026 sets and old-syllabus MCQ 7.Go to answer
- Prove v_f=v_i+at, S=v_it+\tfrac12at^2 and 2aS=v_f^2-v_i^2 by integration. (long)A standard proof (Exercise 5.2 Q.13) behind every constant-acceleration problem.Go to answer
- Show that the slope of an s-t graph is velocity, the slope of a v-t graph is acceleration, and the area under a v-t graph is displacement. (short)The three graph facts of Section 5.2 used in every graph question.Go to answer
- v-t graph problems: acceleration, deceleration and displacement from areas. (problem)Exercise 5.1 Q.9 to Q.14 and Review Q.3 all use this method.Go to answer
- Is the particle speeding up or slowing down? Compare the signs of v and a. (short)Exercise 5.2 Q.4 and Review Q.4 and Q.5 ask it directly.Go to answer
- Vertical projection: time to the top, maximum height, time of flight and impact velocity. (problem)Book Example 5 and Exercise 5.2 Q.1 and Q.3.Go to answer