Kinematics - Cambridge University Press · P1: FXS/ABE P2: FXS 9780521740494c19.xml CUAU033-EVANS September 10, 2008 7:6 Chapter 19 — Kinematics 467 Example 3 A particle moves in

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9780521740494c19.xml CUAU033-EVANS September 10, 2008 7:6

C H A P T E R

19Kinematics

ObjectivesTo model motion in a straight line and to use calculus to solve problems involving

motion in a line with constant and variable acceleration

To use graphical methods to solve problems involving motion in a straight line

IntroductionKinematics is the study of motion without reference to the cause of the motion. In this chapter

we will consider the motion of a particle in a straight line only. Such motion is called

rectilinear motion. When referring to the motion of a particle we may in fact be referring to a

body of any size. However for the purposes of studying its motion we can consider that all

forces that act upon the body, causing it to move, act through a single point. Hence we can

consider the motion of a car or train in the same way as we would consider the motion of a

dimensionless particle.

It is important to make a distinction between vector and scalar quantities when studying

motion. Quantities such as displacement, velocity and acceleration must be specified by both

magnitude and direction. They are vector quantities. Distance, speed and time on the other

hand are specified by their magnitude only. They are scalar quantities.

Since we are considering movement in a straight line only, the direction of all vector

quantities is simply specified by the sign of the numerical value.

19.1 Position, velocity and accelerationPositionThe position coordinate of a particle moving in a straight line is determined by its distance

from a fixed point O on the line, called the origin, and whether it is to the right or left of O.

Conventionally the direction to the right of the origin is considered to be positive.

O P

x

X

463

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464 Essential Advanced General Mathematics

Consider a particle which starts at O and begins to move. The position of a particle is

determined by a number, x, called the position coordinate. If the units are metres and if

x = −3, the position is 3 m to the left of O, while if x = 3, the displacement is 3 m to the

right of O.

The displacement is defined as the change in position of the particle relative to O.

Sometimes there is a rule which enables the position coordinate, at any instant, to be

calculated. In this case x is redefined as a function of t. Hence x(t) is the displacement at

time t. Specification of a displacement function together with the physical idealisation of

a real situation constitute a mathematical model of the situation.

An example of a mathematical model is the following.

A stone is dropped from the top of a vertical cliff 45 m high. Assume that the stone is a

particle travelling in a straight line. Let x(t) be the downwards position of the particle from O,

the top of the cliff, t seconds after the particle is dropped. If air resistance is neglected, an

approximate model for the displacement is

x(t) = 5t2 for 0 ≤ t ≤ 3

It is important to distinguish between the scalar quantity distance and the vector quantity

displacement.

Consider a particle that starts at O and moves firstly five units to the right to point P, and

then seven units to the left to point Q.

1 2 3 4 5 60–1–2–3–4

Q O P

Its final position is x = −2. However the distance the particle has moved is 12 units.

Example 1

A particle moves in a straight line so that its position x cm relative to O at time t seconds is

given by x = t2 − 7t + 6, t ≥ 0. Find

a its initial position b its position at t = 4.

Solution

a At t = 0, x = +6 i.e. the particle is 6 cm to the right of O.

b At t = 4, x = (4)2 − 7(4) + 6 = −6 i.e. the particle is 6 cm to the left of O.

VelocityYou should already be familiar with the concept of a rate of change through your studies in

Mathematical Methods.

The velocity of a particle is defined as the rate of change of its position with respect to time.

We can consider the average rate of change, the change in position over a period of time, or

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Chapter 19 — Kinematics 465

we can consider the instantaneous rate of change, which specifies the rate of change at a

given instant in time.

If a particle moves from x1 at time t1 to x2 at time t2, then its

average velocity = x2 − x1

t2 − t1

Velocity may be positive, negative or zero. If the velocity is positive the particle is moving to

the right and if it is negative the direction of motion is to the left. A velocity of zero means the

particle is instantaneously at rest.

The instantaneous rate of change of position with respect to time is the instantaneous

velocity. If the position, x, of the particle at time t is given as a function of t, then the velocity of

the particle at time t is determined by differentiating the rule for position with respect to time.

Common units of velocity (and speed) are:

1 metre per second = 1 m/s

1 centimetre per second = 1 cm/s

1 kilometre per hour = 1 km/h

The first and third are connected in the following way:

1 km/h = 1000 m/h

= 1000

60 × 60m/s

= 5

18m/s

∴ 1 m/s = 18

5km/h

Note the distinction between velocity and speed.

Speed is the magnitude of the velocity.

Average speed for a time interval [t1, t2] is equal todistances travelled

t2 − t1

Instantaneous velocity v = dx

dtwhere x is a function of time.

Example 2


given by x = t2 − 7t + 6, t ≥ 0. Find

a its initial velocity b when and where its velocity equals zero

c its average velocity for the first 4 s d its average speed for the first 4 s.

Solution

a x = t2 − 7t + 6

v = dx

dt= 2t − 7

at t = 0, v = −7 i.e. the particle is moving to the left at 7 cm/s.

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b 2t − 7 = 0

implies t = 3.5

When t = 3.5, x = (3.5)2 − 7(3.5) + 6

= −6.25

The particle is 6.25 cm to the left of O.

c average velocity = change in position

change in timeat t = 4, x = −6

∴ average velocity = −6 − +6

4= −3 cm/s

d average speed = distance travelled

change in time

1 2 3 4 5 60–1–2–3–4–5–6

O

t = 4t = 0t = 3.5

14

–6

Since the particle has stopped at t = 3.5 and begun to move in the opposite direction,

we must consider the distance travelled in the first 3.5 s (from x = 6 to x = −6.25)

and then the distance travelled in the final 0.5 s (from x = −6.25 to x = −6).

total distance travelled = 12.25 + 0.25 = 12.5

average speed = 12.5

4= 3.125 cm/s

AccelerationThe acceleration of a particle is defined as the rate of change of its velocity with respect to time.

Average acceleration for the time interval [t1, t2] is defined byv2 − v1

t2 − t1where v2 is the

velocity at time t2 and v1 is the velocity at time t1.

Instantaneous acceleration a = dv

dt= d

dt

(dx

dt

)= d2x

dt2

For kinematics, the second derivatived2x

dt2is denoted by x ′′(t) or x (t).

Acceleration may be positive, negative or zero. Zero acceleration means the particle is

moving at a constant velocity. Note that the direction of motion and the acceleration need not

coincide. For example, a particle may have a positive velocity indicating it is moving to the

right, but a negative acceleration indicating it is slowing down. Also, although a particle may

be instantaneously at rest its acceleration at that instant need not be zero. If acceleration has

the same sign as velocity then the particle is ‘speeding up’. If the sign is opposite the particle is

‘slowing down’.

The most commonly used units for acceleration include cm/s2 and m/s2.

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Example 3


given by x = t3 − 6t2 + 5, t ≥ 0. Find

a its initial position, velocity and acceleration and hence describe its motion

b the times when it is instantaneously at rest and its position and acceleration at those times.

Solution

a for x = t3 − 6t2 + 5, v = 3t2 − 12t and a = 6t − 12

t = 0

x = 5, v = 0 and a = −12

Particle is instantaneously at rest 5 cm to right of O with an acceleration of

−12 cm/s2.

b v = 3t2 − 12t = 0

3t(t − 4) = 0

t = 0 or t = 4

Particle is initially at rest and stops again after 4 s.

At t = 0, x = 5 and a = −12

At t = 4, x = (4)3 − 6(4)2 + 5 = −27 and a = 6(4) − 12 = 12

After 4 s the position of the particle is 27 cm to the left of O and its acceleration is

12 cm/s.

Exercise 19A

1 A particle moves in a straight line so that its position x cm relative to O at time t secondsExamples 1, 2

(t ≥ 0) is given by x = t2 − 7t + 12. Find

a its initial position b its position at t = 5

c its initial velocity d when and where its velocity equals zero

e its average velocity in the first 5 s f its average speed in the first 5 s.

2 The position x metres at time t seconds (t ≥ 0) of a particle moving in a straight line isExample 3

given by x = t2 − 7t + 10. Find

a when its velocity equals zero b its acceleration at this time

c the distance travelled in the first 5 s d when and where its velocity is −2 m/s.

3 A particle moving in a straight line is x cm from the point O at time t seconds (t ≥ 0) where

x = t3 − 11t2 + 24t − 3. Find

a its initial position and velocity b its velocity at any time

c at what times the particle is stationary d where the particle is stationary

e for how long the particle’s velocity is negative f its acceleration at any time

g when the particle’s acceleration is zero and its velocity and position at that time.

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4 A particle moves in a straight line so that its position x cm relative to O at time t seconds

(t ≥ 0) is given by x = 2t3 − 5t2 + 4t − 5. Find

a when its velocity is zero and its acceleration at that time

b when its acceleration is zero and its velocity at that time.

5 A particle moving in a straight line is x cm from the point O at time t seconds (t ≥ 0) where

x = t3 − 13t2 + 46t − 48.

Find when it passes through O and its velocity and acceleration at those times.

6 Two particles are moving along a straight path so that their displacements, x cm from a

fixed point P at any time, are given by x = t + 2 and x = t2 − 2t − 2. Find

a the time when the particles are at the same position

b the time when they are moving with the same velocity.

19.2 Using antiderivatives for kinematics problemsSo far we have considered examples where the equation of motion has defined the position of

the particle in terms of time and from it we have derived equations for the velocity and the

acceleration by differentiation.

We may be given a rule for acceleration at time t, and by the use of antidifferentiation with

respect to t and some additional information we can deduce rules for both velocity and

position.

Example 4

A body starts from O and moves in a straight line. After t seconds (t ≥ 0) its velocity (v cm/s)

is given by v = 2t − 4. Find

a its position x in terms of t b its position after 3 s

c its average velocity in the first 3 s d the distance travelled in the first 3 s

e its average speed in the first 3 s.

Solution

a Antidifferentiate with respect to t to find the expression for position x m at time

t seconds

x = t2 − 4t + c

When t = 0, x = 0 and therefore c = 0

x = t2 − 4t

b When t = 3, x = −3. The body is 3 units to the left of O

c Average velocity = −3 − 0

3= −1 m/s

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d v = 0 when 2t − 4 = 0, i.e., when t = 2

When t = 2, x = −4

Therefore the body goes from x = 0 to x = −4 in the first 2 s, then back

to −3 in the next second. It has travelled 5 m in the first 3 s.

e Its average speed is5

3m/s.

Example 5

A particle starts from rest 3 m from a fixed point and moves in a straight line with an

acceleration of a = 6t + 8. Find its position and velocity at any time t seconds.

Solution

a = dv

dt= 6t + 8

by antidifferentiating v = 3t2 + 8t + c

at t = 0, v = 0 and so c = 0

∴ v = 3t2 + 8t

by antidifferentiating again x = t3 + 4t2 + d

at t = 0, x = 3 and so d = 3

∴ x = t3 + 4t2 + 3

Example 6

A stone is projected vertically upward from the top of a building 20 m high with an initial

velocity of 15 m/s.

Find

a the time taken for the stone to reach its maximum height

b the maximum height reached by the stone

c the time taken for the stone to reach the ground

d the velocity of the stone as it hits the ground.

In this case we only consider the stone’s motion in a vertical direction so we can consider it as

rectilinear motion. Also we will assume that the acceleration due to gravity is approximately

−10 m/s2 (note that downward is considered the negative direction).

Solution

Given that a = −10

∴ v = −10t + c

at t = 0, v = 15

∴ v = −10t + 15

∴ x = −5t2 + 15t + d

at t = 0, x = 20

∴ x = −5t2 + 15t + 20

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a The stone will reach its maximum height when v = 0

∴ −10t + 15 = 0

which implies t = 1.5

b At t = 1.5, x = −5(1.5)2 + 15(1.5) + 20

= 31.25

The maximum height reached by the stone is 31.25 m.

c The stone reaches the ground when x = 0

−5t2 + 15t + 20 = 0

−5(t2 − 3t − 4) = 0

−5(t − 4)(t + 1) = 0

∴ t = 4 (solution of t = −1 is rejected since t ≥ 0)

i.e. the stone takes 4 s to reach the ground.

d At t = 4, v = −10(4) + 15

= −25i.e. velocity on impact is −25 m/s.

Exercise 19B

1 A body starts from O and moves in a straight line. After t seconds (t ≥ 0) its velocityExample 4

(v cm/s) is given by v = 4t − 6. Find

a its position x in terms of t b its position after 3 s

c the distance travelled in the first 3 s d its average velocity in the first 3 s

e its average speed in the first 3 s.

2 The velocity (v m/s) at time t seconds (t ≥ 0) of a particle is given by v = 3t2 − 8t + 5. It

is initially 4 m to the right of a point O. Find

a its displacement and acceleration at any time

b its displacement when the velocity is zero

c its acceleration when the velocity is zero.

3 A body moves in a straight line with an acceleration of 10 m/s2. If after 2 s it passesExample 6

through O and after 3 s it is 25 m from O, find its initial displacement relative to O.

4 A body moves in a straight line so that its acceleration(a m/s2) after time t seconds (t ≥ 0)

is given by a = 2t − 3. If the initial position of the body is 2 m to the right of a point O and

its velocity is 3 m/s, find the particle’s position and velocity after 10 s.

5 A body is projected vertically upwards with a velocity of 25 m/s. (Its acceleration due toExample 5

gravity is −10 m/s2.) Find

a the particle’s velocity at any time

b its height above the point of projection at any time

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c the time it takes to reach its maximum height

d the maximum height reached

e the time taken to return to the point of projection.

6 In a tall building the lift passes the 50th floor with a velocity of −8 m/s and an acceleration

of1

9(t − 5) m/s2. If each floor spans a distance of 6 m, find at which floor the lift

will stop.

19.3 Constant accelerationWhen considering motion of a particle due to a constant force, e.g. gravity, the acceleration is

constant. There are a number of rules that we may establish by considering the case where

acceleration remains constant or uniform.

Given thatdv

dt= a

by antidifferentiating we have

v = at + c where c is the initial velocity.

Using the symbol u for initial velocity we have

v = u + at 1

Now given thatdx

dt= v

by antidifferentiating a second time we have

x = ut + 1

2at2 + d, where d is the initial position.

If we consider s = x − d as the change in position of the particle from its starting point,

i.e. the particle’s displacement from its initial position, we have

v = ut + 1

2at2 2

If we transform the formula v = u + at so that t is the subject we have

t = v − u

a

By substitution in s = ut + 1

2at2

s = u(v − u)

a+ a(v − u)2

2a2

2as = 2u(v − u) + (v − u)2

= 2uv − 2u2 + v2 − 2uv + u2

= v2 − u2

i.e. v2 = u2 + 2as 3

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Also we know that distance travelled = average velocity × time.

i.e. s = 1

2(u + v)t 4

These four formulas are very useful but it must be remembered that they only apply when

dealing with constant acceleration.

When approaching problems involving constant acceleration it is a good idea to list the

quantities you are given, establish which quantity or quantities you require and then use the

appropriate formula. Ensure that all quantities are converted to compatible units.

Constant acceleration summaryIf acceleration is constant, the following formulas may be applied, where u is the initial

velocity, v is the final velocity, a is the acceleration, t is the time and s is the displacement.

v = u + at s = ut + 1

2at2 v2 = u2 + 2as s = 1

2(u + v)t

Example 7

A body is moving in a straight line with uniform acceleration at an initial velocity of 12 m/s.

After 5 s its velocity is 20 m/s. Find

a the acceleration b the distance travelled in this time

c the time taken to travel a distance of 200 m.

Solution

Given u = 12

v = 20

t = 5

a Find a using v = u + at

20 = 12 + 5a

a = 1.6The acceleration is 1.6 m/s2.

b Find s using s = ut + 1

2at2

= 12(5) + 1

2(1.6)52

= 80

The distance travelled is 80 m.

c Using the formula s = ut + 1

2at2 gives

200 = 12t + 1

2× (1.6) × t2

200 = 12t + 4

5t2

1000 = 60t + 4t2

250 = 15t + t2

i.e. t2 + 15t − 250 = 0

(t − 10)(t + 25) = 0

∴ t = 10 or t = −25

As t ≥ 0, t = 10 is the acceptable solution.

The body takes 10 s to travel a distance of 200 m.

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Exercise 19C

1 How long does it take for a body at rest to travel a distance of 30 m if it is accelerated at

1.5 m/s2?

2 A car is travelling at 25 m/s when the brakes are applied. It is brought to rest with uniform

deceleration in 3 s. How far would it travel after the brakes were applied?

3 A motor cycle accelerates uniformly from 3 m/s to 30 m/s in 9 s. FindExample 7

a the acceleration

b the time it will take to increase in speed from 30 m/s to 50 m/s

c the distance travelled in the first 15 s (assuming it starts from rest)

d the time taken to reach a speed of 200 km/h (assuming it starts from rest).

4 A car accelerating uniformly from rest reaches a speed of 45 km/h in 5 s. Find

a its acceleration

b the distance travelled in the 5 s.

5 A train starts from rest at a station and accelerates uniformly at 0.5 m/s2 until it reaches a

speed of 90 km/h.

a How long does the train take to reach this speed?

b How far does the train travel in reaching this speed?

6 A train travelling at 54 km/h begins to climb an incline of constant gradient that produces

a deceleration of 0.25 m/s2.

a How long will the train take to travel a distance of 250 m?

b What will the train’s speed be then?

For 7 to 11 assume that the acceleration due to gravity is − 9.8 m/s2 and ignore air

resistance. Upward motion is considered to be in the positive direction.

7 A stone is projected vertically upwards from O with a speed of 20 m/s. Find

a the velocity of the stone after 4 s

b the distance of the stone from O after 4 s.

8 Repeat 7 for the stone being projected downwards from O with the same speed.

9 A body is projected vertically upwards with a velocity of 49 m/s.

a After what time will the body return to the point of projection?

b When will the body be at a height of 102.9 m above the point of projection?

10 A man dives from a springboard where his centre of gravity is initially 3 m above the

water and his initial velocity is 4.9 m/s upwards. Regarding the diver as a particle at his

centre of gravity, and assuming that the diver’s motion is vertical, find

a the diver’s velocity after t seconds

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b the diver’s height above the water after t seconds

c the maximum height of the diver above the water

d the time taken for the diver to reach the water.

11 A stone is thrown vertically upwards from the top edge of a cliff 24.5 m high with a speed

of 19.6 m/s. Find


b the maximum height reached from the foot of the cliff

c the time taken for the stone to return to the point of projection

d the time taken for the stone to reach the foot of the cliff.

12 A body is travelling at 20 m/s when it passes point P and 40 m/s when it passes point Q.

Find its speed when it is halfway from P to Q, assuming uniform acceleration.

19.4 Velocity–time graphsMany kinematics problems can be solved using velocity–time graphs. These are particularly

useful if acceleration is constant but with a broader knowledge of integral calculus they can

also be used when acceleration is variable.

First, we understand that if the acceleration is constant then v = u + at .

This constitutes a linear relationship between v and t where a is the gradient of the

corresponding velocity–time graph.

Since v = dx

dtit follows that

∫ t2

t1

v(t)dt = x2 − x1

where x1 is the position at time t1 and x2 is the position at time t2.

Then the total area of the region(s) between

the velocity–time graph and the t axis

corresponds to the distance travelled by

the particle between times t1 and t2. area =displacement

v

tt2t1

Consideration of the velocity–time graph

is particularly useful in situations where

there are several stages to the particle’s

motion.

Example 8

A car starts from rest and accelerates uniformly for 25 s until it is travelling at 25 m/s. It

then maintains this velocity for 3 minutes before decelerating uniformly until it stops in

another 15 s.

Construct a velocity–time graph and use it to determine the total distance travelled in

kilometres.

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Solution

From the graph we can calculate

the area of the trapezium.

Area = (a + b) h

2

= 1

2(220 + 180) 25

= 5000 m

= 5 km

The total distance travelled is 5 km.

25

25 205 220O t (s)

v (m/s)

Example 9

A motorist is travelling at a constant speed of 120 km/h when he passes a stationary police car.

He continues at that speed for another 15 s before uniformly decelerating to 100 km/h in 5 s.

The police car takes off after the motorist the instant it passes. It accelerates uniformly for 25 s

by which time it has reached 130 km/h. It continues at that speed until it catches up to the

motorist. After how long does the police car catch up to the motorist and how far has he

travelled in that time?

Solution

We start by representing the information

on a velocity–time graph.130

100

2015 25

120

t (s)

v (km/h)

police car

motorist

TO

The distance travelled by the motorist

and the police car will be the same so

the areas under each of the velocity–time

graphs will be equal. This fact can be

used to find T, the time taken for the

police car to catch up to the motorist.

For the motorist, the distance travelled

after T seconds

=(

120 × 15 + 1

2(120 + 100) × 5 + 100(T − 20)

)5

18

= (1800 + 550 + 100T − 2000)5

18

= (100T + 350)5

18

Note: The factor5

18changes velocities from km/h to m/s.

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Police car:

(1

2× 25 × 130 + 130(T − 25)

)5

18

= (1625 + 130T − 3250)5

18

= (130T − 1625)5

18

When the police car catches the motorist

100T + 350 = 130T − 1625

30T = 1975

T = 395

6

The police car catches the motorist after 65.83 s.

∴ distance = (100T + 350)5

18where T = 395

6

= 52 000

27m

distance = 1.926 km

The police car has travelled 1.926 km when it catches the motorist.

Exercise 19D

It is suggested that you draw a velocity–time graph for each of these questions.

1 A particle starts from rest and accelerates uniformly for 5 s until it reaches a speed of

10 m/s. It immediately decelerates uniformly until it comes to rest after a further 8 s. How

far did it travel?

Example 8

2 A car accelerates uniformly from rest for 10 s to a speed of 15 m/s. It maintains this speed

for a further 25 s before decelerating uniformly to rest after a further 15 s. Find

a the total distance travelled by the car

b the distance it had travelled when it started to decelerate

c the time taken for it to reach the halfway point of its journey.

3 A particle starts from rest and travels 1 km before coming to rest again. For the first 5 s it

accelerates uniformly. It next maintains a constant speed for 500 m, and then decelerates

uniformly for the last 10 s. Find the maximum speed of the particle.

4 A car passes point P with a speed of 36 km/h and continues at this speed for 12 s before

accelerating to a speed of 72 km/h in 6 s. How far from P is the car when it reaches a speed

of 72 km/h?

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5 A tram decelerates uniformly from a speed of 60 km/h to rest in 60 s. Find

a the distance travelled by the tram

b how far it had travelled by the time it had reduced its speed by half

c the time taken for it to travel half the total distance.

6 A car passes a point A with a speed of 15 m/s and continues travelling at that speed. AExample 9

second car starts from rest and accelerates uniformly until it reaches a speed of 25 m/s in

10 s. Both cars continue with a constant speed on to point B which they reach at the same

time.

a How long does it take for both cars to reach point B?

b How far is it from A to B?

7 Two stations A and B are 14 km apart. A train passes through station A, heading towards B,

maintaining a constant speed of 60 km/h. At the instant it passes through A, a second train

on the same track leaves station B, heading towards A, and accelerates uniformly. After 5

minutes the alarm is raised at both stations simultaneously that a collision is imminent.

Both trains are radioed and told to brake. The first train decelerates uniformly so that it will

stop in 2.5 minutes. The second train, which has reached a speed of 80 km/h, will take 4

minutes to stop. Will they collide?

8 Two tram stops are 800 m apart. A tram starts at rest from the first stop and accelerates at a

constant acceleration of a m/s2 for a certain time and then decelerates at a constant rate of

2a m/s2, before coming to rest at the second stop. The time taken to travel between the

stops is 1 min 40 seconds. Find

a the maximum speed reached by the tram in km/h

b the time at which the brakes are applied

c the value of a.

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Rev

iew


Chapter summary

The position coordinate of a particle moving in a straight line is determined by its distance

from a fixed point O, called the origin, and whether the particle is to the right or left of O.

Conventionally, the direction to the right of the origin is considered to be positive.� displacement (x) is the position of the particle relative to O� velocity (v) is the rate of change of its position with respect to time, i.e. v = dx

dt� speed is a scalar quantity and refers to the distance travelled per unit time

� average velocity = change in position

change in time

� average speed = distance travelled

change in time� acceleration (a) is the rate of change of its velocity with respect to time, i.e.

a = dv

dt= d2x

dt2

Constant acceleration

If acceleration is constant, the following formulae may be applied where u is the initial

velocity, v is the final velocity, a is the acceleration, t is the time and s is the displacement� v = u + at � s = ut + 1

2at2

� v2 = u2 + 2as � s = 1

2(u + v) t

Velocity–time graphs

The area of the region(s) between the velocity–time (v against t) graph and the t axis

between t = t1 and t = t2 corresponds to the distance travelled by the particle between

times t1 and t2.

Multiple-choice questions

1 A particle moves in a straight line so that its position x cm from a fixed point O at time t

seconds (t ≥ 0) is given by x = −t3 + 7t2 − 12t . The initial position of the particle

relative to O is

A 0 cm B −6 cm C 12 cm D −20 cm E 5 cm

2 A particle moves in a straight line so that its position x cm from a fixed point O at time t

seconds (t ≥ 0) is given by x = −t3 + 7t2 − 12t . The average velocity of the particle in

the first 2 s correct to two decimal places is

A 4 cm/s B −4 cm/s C 2 cm/s D 4.06 cm/s E −2 cm/s

3 A particle moves in a straight line with acceleration of 4 − 6t m/s2 at time t seconds. The

particle has an initial velocity of −1 m/s and an initial position of 4 m from a fixed point O.

The velocity of the particle when t = 1 is

A −1 m/s B 6 m/s C 0 m/s D 4 m/s E −2 m/s

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4 A body starts from rest with a uniform acceleration of 1.8 m/s2. The time it will take for

the body to travel 90 m is

A 5 s B√

10 s C 10 s D√

10 E 10√

2 s

5 A car accelerating uniformly from rest reaches a speed of 60 km/h in 4 s. The car’s

acceleration is

A 15 km/h2 B 15 m/s2 C 54 m/s2 D25

6km/h2 E

25

6m/s2

6 A car accelerating uniformly from rest reaches a speed of 60 km/h in 4 s. The distance

travelled by the car in the 4 s is

A 200 m B 100 km C100

3m D 100 m E 360 m

7 A car’s motion is represented by the

velocity–time graph shown.

10

25

20

15

15

10

t (s)

v (m/s)

5 60

5

4

The total distance travelled by the car

over the 15 s is

A 75 m B 315 m C 182.5 m

D 167.5 m E 375 m

8 A rock falls from the top of a cliff 40 m high. The rock’s speed just before it hits the ground

in m/s(g = 9.8 m/s2) is

A 20 B 22 C 24 D 26 E 28

9 A body initially travelling at 20 m/s is subject to a constant deceleration of 4 m/s2. The

time it takes to come to rest (t seconds) and the distance travelled before it comes to rest

(s metres) is given by

A t = 5, s = 50 B t = 5, s = 45 C t = 4, s = 20

D t = 5, s = 40 E t = 4, s = 35

10 A particle moves in a straight line with acceleration of 12t − 5 m/s2 at time t seconds. The

particle has an initial velocity of 1 m/s and an initial position of 0 m from a fixed point O.

Find the velocity of the particle at t = 1.

A 1 m/s B −5 m/s C 7 m/s D 2 m/s E 3 m/s

Short-answer questions (technology-free)


(t ≥ 0) is given by x = t2 − 4t − 5. Find

a its initial position b its position at t = 3

c its initial velocity d when and where its velocity equals zero

e its average velocity in the first 3 s f its average speed in the first 3 s.


(t ≥ 0) is given by x = t3 − 2t2 + 8. Find

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a its initial position, velocity and acceleration and hence describe its motion

b the times when it is stationary and its positions and acceleration at those times.

3 A particle moving in a straight line is x cm from the point O at time t seconds (t ≥ 0),

where x = −2t3 + 3t2 + 12t + 7. Find

a when the particle passes through O and its velocity and its acceleration at those times

b when the particle is at rest

c the distance travelled in the first 3 s.

4 Two particles A and B are moving in a straight line such that their displacements x cm from

the point O at time t seconds are given by x1 and x2 respectively, where

x1(t) =t3 − t2 t ≥0

x2(t) =t2 t ≥0

a Find

i the displacement of A after1

2s ii the acceleration of A after

1

2s

iii the velocity of B after1

2s.

b Find

i the times when A and B collide (i.e., have the same displacement)

ii the maximum distance between A and B during the first 2 s of motion.

5 A particle moving in a straight line has acceleration of 6t m/s2 at time t seconds (t ≥ 0). If

the particle starts from rest at the origin O, find

a the velocity after 2 s b the displacement at any time t.

6 A particle moving in a straight line has acceleration of (3 − 2t) m/s2 at time t seconds

(t ≥ 0). If the particle starts at the origin O with a velocity of 4 m/s, find

a the time when the particle comes to rest

b the position of the particle at the instant it comes to rest

c the acceleration at this instant

d the time when the acceleration is zero

e the velocity at this time.

7 A particle moves in a straight line and, t seconds after it starts from O, its velocity is(2t2 − 3t3

)m/s. Find

a the displacement after 1 s

b the velocity after 1 s

c the acceleration after 1 s.

8 For a particle moving in a straight line, the velocity function is v : R+ → R, where

v (t) = 1

2t2and t is the time in seconds. Find

a the acceleration after t seconds

b the displacement at time t seconds, given that the particle is at O when t = 1.

9 The velocity, v m/s, of a body t seconds after it starts moving from O along a straight line is

given by v = t3 − 11t2 + 24t, t ≥ 0.

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a Find the acceleration at time t.

b Find the acceleration at the instant when the body first changes direction.

c Find the displacement of the body from O after 5 s, and the total distance travelled in

the first 5 s.

10 A car is travelling at 20 m/s when the brakes are applied. It is brought to rest with uniform

deceleration in 4 s. How far did it travel after the brakes were applied?

11 A car accelerates uniformly from 0 to 30 m/s in 12 s. Find

a the acceleration

b the time it will take to increase in speed from 30 m/s to 50 m/s

c the distance travelled in the first 20 s

d the time taken to reach a speed of 100 km/h.

12 A train starts from rest at a station and accelerates uniformly at 0.4 m/s2 until it reaches a

speed of 60 km/h.

a How long does the train take to reach this speed?

b How far does the train travel in reaching this speed?

For questions 13 and 14 assume that the acceleration due to gravity is −9.8 m/s2 and ignore air

resistance. Upward motion is considered to be in the positive direction.

13 A body is projected vertically upward with a velocity of 35 m/s.

a After what time will the body return to the point of projection?

b When will the body be at a height of 60 m above the point of projection?

14 A stone is projected vertically upward from the top of a cliff 20 m high with a speed of

19.6 m/s. Find


b the maximum height reached, with respect to ground level

c the time taken for the stone to return to the point of projection

d the time taken for the stone to reach the foot of the cliff.

It is suggested that you draw a velocity–time graph for each of the questions 15 to 18.

15 A particle starts from rest and accelerates uniformly for 15 s until it reaches a speed of

25 m/s. It immediately decelerates uniformly until it comes to rest after a further 20 s. How

far did it travel?

16 A car accelerates uniformly from rest for 8 s to a speed of 12 m/s. It maintains this speed

for a further 15 s before decelerating uniformly to rest after a further 10 s. Find

a the total distance travelled by the car

b the time taken for it to reach the halfway point of its journey.

17 A vehicle starts from rest and travels 1 km before coming to rest again. For the first 15 s it

accelerates uniformly, before maintaining a constant speed for 800 m then finally

decelerating uniformly to rest in 10 s. Find the maximum speed of the vehicle.

18 A car travels at a constant speed of 12 m/s along a straight road. It passes a second

stationary car which sets off in pursuit 3 s later. Find the constant acceleration required for

the second car so that it catches the first car after a further 27 s has passed.

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19 A particle moves in a straight line so that t seconds after passing a fixed point O in the line,

its velocity, v metres per second, is given by v = t2

4− 3t + 5.

Calculate:

a the velocity after 10 s b the acceleration when t = 0 c the minimum velocity

d the distance travelled in the first 2 s e the distance travelled in the 3rd second.

20 A spot of light moves along a straight line so that its acceleration t seconds after passing a

fixed point O on the line is (2 − 2t) cm/s2. Three seconds after passing O the spot has a

velocity of 5 cm/s. Find, in terms of t, an expression for

a the velocity of the spot of light after t seconds

b the distance of the spot from O after t seconds.

21 A point P is moving along a straight line. It passes through a point O with a velocity 6 m/s

and, t seconds after passing through O, its acceleration is (4 − 4t) m/s2.

a Show that, t seconds after passing through O, the velocity of P is(6 + 4t − 2t2

)m/s.

b Calculate

i the maximum velocity of P

ii the value of t when the velocity of P is again 6 m/s

iii the distance OP when the velocity of P is zero.

22 A particle travelling in a straight line passes a fixed point O with a velocity 5 m/s. Its

acceleration, a m/s2, is given by a = 27 − 4t2, where t seconds is the time after passing O.

Calculate

a the acceleration of the particle as is passes O

b its velocity when t = 3

c the value of t when its velocity is again 5 m/s.

23 A particle passes a fixed point O with a velocity of 2 m/s and moves in a straight line with

acceleration of 3 (1 − t) m/s2, where t is the time in seconds after passing O. Calculate

a the velocity when t = 4 b the position of the particle at this instant.

24 A particle P travels in a straight line from a fixed point O so that its velocity, v m/s, is given

by v = t2 − 10t + 24, where t is the time in seconds after leaving O. Find

a the values of t for which P is instantaneously at rest

b the distance OP when t = 3

c the range of values of t for which the acceleration is negative.

Extended-response questions

1 A particle moves in a straight line so that its displacement x cm relative to O at time

t seconds is given by x = 1

3t3 − 2t2 + 4t − 2

1

3. Find

a its initial displacement b its initial velocity

c its acceleration after 3 s d the time when its velocity equals zero

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e the displacement when its velocity equals zero

f the time when its displacement is zero.

2 A particle moves in a straight line so that its displacement, x cm, relative to O at time

t seconds (t ≥ 0) is given by x = t4 + 2t2 − 8t . Show that

a the particle moves first to the left

b the greatest distance of the particle to the left of O occurs after 1 s

c after this time, the particle always moves to the right.

3 A defective rocket rises vertically upwards into the air and then crashes back to the ground.

The rocket’s height above the ground, at time t seconds after take-off, is h metres where

h = 6t2 − t3 (an approximate model).

a When does the rocket crash and what is its velocity at this time?

b At what time is the speed of the rocket zero, and what is its maximum height?

c When does the acceleration of the rocket become negative?

4 A body is projected vertically upwards at 20 m/s from the top of a tower 10 m high on the

edge of a vertical cliff. The upward displacement, x(t) metres, of the body from ground

level O at time t seconds after projection (t ≥ 0) is given by x (t) = −4.9 t2 + 20t + 10.

Use a graphics calculator or otherwise to evaluate the values

x (1) − x (0) , x (2) − x (1) , x (3) − x (2) , . . . , x (10) − x (9) .

Analyse your results and draw some inference about the motion of the body.

5 A particle is projected vertically upwards with a speed of u m/s. Prove that

a the time taken by the particle to reach its highest point isu

gseconds

b the total time taken for the particle to return to the point of projection is2u

gseconds

c the speed of returning to the point of projection is u m/s.

6 A stone is projected vertically upwards with a speed of 14 m/s, from a point O at the top of

a mine shaft. Five seconds earlier, a lift began to descend the mine shaft from O with a

constant speed of 3.5 m/s. Find the depth of the lift (to the nearest metre) at the instant

when the stone falls on it. (Neglect air resistance and take the acceleration due to gravity as

9.8 m/s2.)

7 A car is travelling along a straight road at 90 km/h when the brakes are applied. The car

comes to rest in 5 s and during this time the velocity decreases linearly with time. Find:

a the rule for the velocity–time function after the brakes are applied

b the distance travelled in the five seconds.

8 A particle moves in a straight line so that its displacement x cm relative to O at time

t seconds (t ≥ 0) is given by x = 3t4 − 4t3 + 24t2 − 48t . Show that the particle moves at

first to the left, comes to rest at a point A and then moves always to the right. Find the

position of A.

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9 A particle is projected vertically upwards with a velocity of u m/s from a point O on the

ground. T seconds later a second particle is projected vertically upwards from O with the

same velocity.

Prove that the time taken for the particles to collide is

(u

g+ T

2

)seconds after the first

particle was launched, and that the height of the particles at the instant they collide is

4u2 − g2T 2

8gmetres above O.

Interpret the case where T = 2u

g. What happens if T >

2u

g?

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Kinematics - Cambridge University Press · P1: FXS/ABE P2: FXS 9780521740494c19.xml CUAU033-EVANS September 10, 2008 7:6 Chapter 19 — Kinematics 467 Example 3 A particle moves in