kinematics - mathsbooks.netmathsbooks.net/jacplus books/12 specialist/ch08 kinematics.pdfchapter 8...

8A Differentiation and displacement, velocity and acceleration

8B Using antidifferentiation 8C Motion under constant acceleration 8D Velocity–time graphs 8E Applying differential equations to

rectilinear motion

8

368

The application of differentiation, • antidifferentiation and solution of differential equations to rectilinear motion of a single

particle, including the different derivative forms for acceleration

ad x

dt

dxdt

vdvdx

ddx

v= == = = == =v= =v

2d x2d x2

212

Velocity–time graphs and their use.• Velocity–time graphs and their use.

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Digital doc10 Quick Questions

areaS oF STUdy

Kinematics

differentiation and displacement, velocity and accelerationintroductionThe motion of stellar and earthly objects — like the sweeping movement of a planet, the curving path of a thrown javelin, the acceleration of a stone dropped into a river and the graceful turnings of a ballet dancer — have intrigued people for centuries. Consequently, the study of the motion of bodies, objects and particles — termed kinematics — has grown until now it is fundamental to most sciences. This chapter will deal with motion in a straight line only — rectilinear motion — and all objects considered will be treated as points for mathematical convenience; that is, the objects do not rotate or change shape. First, some basic concepts are reviewed and explained.

positionThe position of a particle moving in a straight line measures its location from a fi xed point of reference, usually the origin O on the line. Positions to the right of O are usually taken as positive. The point P in this diagram has a position coordinate x.

displacementThe displacement of a particle is its change in position relative to a fi xed point. Displacement gives both the distance and direction that a particle is from a fi xed point.

8a

Positivedirection

O P

xx

−2 0

F AS4 x

maths Quest 12 Specialist mathematics for the Casio Classpad

369Chapter 8 Kinematics

For example, a particle that moves from S to F via A is shown here on a position–time line.The distance travelled from S to F is 10 units. The displacement of S to F is −2.

VelocityThe average velocity of a particle is the rate of change of its position with respect to time. This can be illustrated on a position–time graph. For this curve the position x at time t is shown. The curve x(t) is shown at right.

Average velocity =

= −−

change in positionchange in time

x xt2 1

2 tt

xt

1

= δδ

The instantaneous velocity, v(t), at time t is defined as the limiting value of the average velocity as δ t approaches zero. That is, it is the gradient of a displacement–time graph

or v txtt

( ) lim=→δ

δδ0

or v tdxdt

( ) =

Units of velocityThe units of velocity are most commonly cm/s, m/s, or km/h.Note: 1 m/s = 3.6 km/h (verify this).

SpeedInstantaneous speed is the magnitude of instantaneous velocity and is always positive; that is, | v(t) |.

Average speeddistance travelled

time taken=

Note that distance travelled is not necessarily the same as displacement.

WorKed example 1

A particle moves in a straight line, left and right of an origin O. The position, x(t), of the particle at any time, t (seconds), is given by:

x(t) = t2 − 2t − 8Assume that the negative direction is to the left of the origin.Find:a the initial position of the particle b the position after 2 secondsc the position after 3 seconds d the average velocity during the third seconde the velocity at any time t f the initial velocityg the velocity after 2 seconds h when and where the particle is stationary.Hence, sketch a position–time line.

ThinK WriTe

a 1 The initial position occurs when t = 0.Find x when t = 0.

a x(t) = t2 − 2t − 8When t = 0, x(0) = 02 − 2(0) − 8 = −8

2 State its position. The initial position is −8 cm, or 8 cm left of 0.

Time

Posi

tion

x2

Change intimeδt

Change inposition

δxx1

t1 t2

370

b 1 Find x when t = 2. b x(2) = 22 − 2(2) − 8= −8

2 State its position. The position after 2 seconds is −8 cm, or 8 cm left of the origin.

c 1 Find x when t = 3. c x(3) = 32 − 2(3) − 8= −5

2 State its position. The position after 3 seconds is −5, or 5 units to the left of the origin.

d Average velocitychange in position

change i=

nn timed Average velocity during the third second

=

=

x x( )x x( )x x( )

( )− −( )− −( )

3 2x x3 2x x−x x−3 2−x x−( )3 2( )x x( )x x3 2x x( )x x( )3 2( )3 2−3 2−5 8−5 8−− −5 8− −( )5 8( )− −( )− −5 8− −( )− −( )5 8( )− −( )− −5 8− −( )− −

1cm

s= 3 cm/s

e Since velocity is the rate of change of position, differentiate x with respect to t.

edxdt

t= −2 2

or v = 2t − 2f Initial velocity when t = 0.

Find v when t = 0.f When t = 0, v(0) = 2(0) − 2

= −2Initial velocity is −2 cm/sor 2 cm/s to the left.

g Find v when t = 2. g When t = 2, v(2) = 2(2) − 2 = 2Velocity after 2 seconds is 2 cm/s to the right.

h 1 The particle is stationary when the velocity is zero. Find t when v = 0.

h The particle is stationary whenv = 2t − 2 = 0.

2 Factorise the LHS. 2(t − 1) = 03 Solve for t. t = 14 Evaluate x when t = 1. x(1) = 12 − 2(1) − 8

= −95 State the answers. The velocity is 0 when t = 1 s and its

displacement from O is −9 cm, or 9 cm to the left of O.

6 Verify using the graph of x(t) on a CAS calculator.

For the displacement–time line:(a) list useful information: position at t = 0, 1, 2 and 3.

v = 0 when t = 1 so the particle turns at t = 1 s

x(0) = −8,x(1) = −9,x(2) = −8,x(3) = −5,v(1) = 0

(b) sketch the position–time line.

0−9 −8 −5 x(cm)

t = 2 st = 1 s t = 0

t = 3 s



WorKed example 2

The displacement, x(t), of a particle moving in a horizontal straight line from an origin O, at any time t is given by:

x t t t t( ) = − − +3 2 2 112

cm

Find:a the initial displacementb when and where the velocity of the particle is 0c the average velocity of the particle during the first 2 seconds of motiond the average speed in the first 2 seconds.

ThinK WriTe

a 1 Find x when t = 0. a x t t t t( ) = − − +3 12

2 2 1

x( ) ( ) ( ) ( )0 0 0 2 0 13 12

2= − − += 1

2 State the initial position. The particle is initially at x = 1or 1 cm right of O.

b 1 Differentiate x to find v. b v tdxdt

( ) =

= 3t2 − t − 2

2 Set v = 0. When v = 0, 3t2 − t − 2 = 0.

3 Factorise the LHS. (3t + 2)(t − 1) = 0

4 Solve for t, giving solutions only where t ≥ 0. t = − 23 or t = 1

Therefore t = 1 is the only solution, since t ≥ 0.

5 Evaluate x(1). x( ) ( ) ( ) ( )1 1 1 2 1 13 12

2

12

= − − +

= −

6 State the answers. The velocity is 0 when t = 1 s and the

position is −1

2 cm or 12 left of O.

7 Verify using a graph of x(t) on a graphics or CAS calculator.

c 1 Evaluate x when t = 2. c x( ) ( ) ( )2 2 2 2 2 1

3

3 12

2= − − +

=

2 Substitute the values into the average velocity rule.

Average velocitychange in position

change i=

nn time

Average velocity = −−

x x( ) ( )2 02 0

= −3 12

3 Evaluate. = 22

= 1

372

4 State the solution. The average velocity in the first 2 seconds is 1 cm/s to the right.

d 1 Sketch a position–time line for the first 2 seconds of motion.

d

O

Origin

−1 1 2 3

t = 1 s t = 2 st = 0

x (cm)

2 Calculate the distance travelled in the first 2 seconds.

Distance travelled = 5 cm

3 Calculate the average speed. Average speed = 52

= 2.5Hence, the average speed for the first 2 seconds is 2.5 cm/s.

Note: The distance travelled, and also the average speed, can be calculated without a line sketch, providing the time(s) when the velocity is found and the appropriate positions found are given.

That is, in part d above: distance = | x(2) − x(1) | + | x(1) − x(0)| = | − | + | − |− −

3 112

12

= | | + | |−3 112

12

= +3 112

12

= 5

accelerationThe average acceleration, a, of a particle during a time interval t2 − t1 is the rate of change of its velocity with respect to time.

That is, average acceleration = v vt t2 1

2 1

−−

= δδ

vt

The instantaneous acceleration, a, at time t is the limiting value of the average acceleration as t approaches 0. That is, it is the gradient of a velocity–time graph, or

a vtt

=→

limδ

δδ0

advdt

=

Units of accelerationThe units of acceleration match their corresponding velocities and are usually expressed as cm/s2 or m/s2.Note: The downward acceleration due to gravity is g = 9.8 m/s2. For objects travelling through the air, air resistance is ignored unless it is specified in a given problem.

Time

Vel

ocity

v2

Changein time

δ t

Change invelocity

δvv

1

t1 t

2



Symbols and unitsThroughout this chapter the following symbols and units are used for kinematic variables.

Symbols Common units

Displacement: x cm m km

Velocity: v or dxdt

or x cms

ms

kmh

Acceleration: a or dvdt

or d x

dt

2

2 or x

cm

s2

m

s2

km

h2

WorKed example 3

The position of a particle moving in a straight line is given by:x(t) = 2t3 + t loge (t) − 4, t > 0.

Find:a the velocity at any time t b the acceleration at any time t.

ThinK WriTe

a 1 a

2

3 Write the solution using the correct notation and variables.

The velocity at any time is given as:v(t) = loge (t) + 6t2 + 1.

To determine the velocity at any time, differentiate x with respect to t, using a CAS calculator by completing the following steps.Defi ne the function x(t). To do this, on the Main screen, tap:• Action• Command• Defi neComplete the entry line as:Defi ne x(t) = 2t3 + t × ln(t) − 4Then press E.

Differentiate x with respect to t. To do this on the soft keyboard, tap:•)•-Complete the entry line as:ddt

x t( (x t( (x t))

Then press E.

374

b 1 b See the screen above.

2 Write the solution using the correctnotation and variables.

The acceleration at any time is given as:

a t tt

( ) = +121

.

WorKed example 4

Find the acceleration in terms of x if x(t) = sin (2t) − cos (2t).

ThinK WriTe

1

2 Write the solution in terms of x. a(t) = 4 cos (2t) − 4 sin (2t)a(t) = −4 (sin (2t) − cos (2t))But x = sin (2t) − cos (2t)a(x) = −4x

Position gives the location of a particle relative to a reference point (usually the 1. origin). The variable used is x.Displacement is change in position 2. δx or s.

Instantaneous velocity, 3. v or dxdt

or x, is the time rate of change of position or the rate of

displacement.

Average velocity (during time interval 4. δ t) = δδ δxt

st

= = displacementtime

.

Speeddistance travelled

time taken=5.

Instantaneous acceleration, 6. a or x or d x

dt

2

2 or

dvdt

, is the time rate of change of velocity.

remember


To determine the acceleration at any time, differentiate the velocity with respect to t. To do this, complete the entry line as:ddt

t t( ( ) )t t) )t t( (In( (t t) )t t+ +t t) )t t6 1) )6 1) )t t) )t t6 1t t) )t t6 1) )6 1) )t t) )t t6 1t t) )t t) )+ +) )6 1) )+ +) )t t) )t t+ +t t) )t t6 1t t) )t t+ +t t) )t t26 126 1) )6 1) )2) )6 1) )) )+ +) )6 1) )+ +) )2) )+ +) )6 1) )+ +) )

Then press E.

To fi nd the acceleration given the position, on the Main screen, complete the entry line as:

d

dtt t

2

22 2t t2 2t t(sin(sin(s ( )t t( )t t2 2( )2 2t t2 2t t( )t t2 2t tt tcot tt t2 2t tcot t2 2t ts (t ts (t t2 2s (2 2t t2 2t ts (t t2 2t t))t t2 2t t−t t2 2t t

Then press E.Note: The second derivative template is located on the soft keyboard, ) and -.


differentiation and displacement, velocity and acceleration 1 Match the sketch of the displacement–time line below to each of the following objects

moving in a straight line, where x is the displacement from O (the origin) at time t and is given by:a x = t2 − 3t + 2 b x = t2 − 2t − 3c x = 2t2 − t − 3 d x = 3t2 − 7t + 2 i

x−4−3−2−1 O 1 2 3 4 5

t = 2t = 1

t = 3 t = 4t = 0

ii

x−2−1 O 1 2 3 4 5 6 7 8

t = 2t = 1

t = 3t = 0

iii

x654321O

t = 2 t = 3 t = 4t = 1 t = 0

ivt = 0

x54321O−3 −2 −1

t = 1 t = 2

2 For each of the following position–time lines, state: i the displacement of F from O ii the displacement of F from S iii the distance travelled to get from S to F.a

2O 1 3 4 5

FS

x

b

654321O−1−2

FS

x

c

21O−1−2−3−4−5−6 x

FS

d

4321O−1−2−3−4−5 x

FS

3 In each case in question 2 it takes 4 seconds to get from S to F. For each case, find:

i the average speed ii the average velocity.

4 mC Use the position–time graph at right.a The average speed in the first 6 seconds is:

A −3 m/s B 5 13 m/s

C 3 m/s D −7 m/s

E 7 m/s

b The average velocity in the first 6 seconds is:

A −3 m/s B 5 13 m/s

C 3 m/s D −7 m/s

E 7 m/s

5 The position–time graph for a particle travelling in a straight line is shown below.Use the graph to:a state the position when i t = 3 ii t = 12b state the time when the displacement is i 24 ii 0c find the average velocity from i t = 3 to t = 7 ii t = 3 to t = 12d find the average speed from i t = 3 to t = 7 ii t = 3 to t = 12.

exerCiSe

8a

Met

res 8

0

(2, 20)

(6, _10)

Seconds

t

x

0

(10, 24)

(12, 18)(7, 10)

(3, −6)5 10

−10

376

6 We 1 A particle moves in a straight line, left and right of an origin O. The position, x(t), of the particle at any time, t (seconds), is given by:

a x(t) = 3t2 − 12 b x(t) = t3 − 12t + 2 c x tt

t( ) =

+4

1

d x tt

t( ) =

+2

1e x(t) = et − 5t − 4 f x(t) = sin (4t)

For the above find: i the initial position ii the position after 3 seconds iii the position after 4 seconds iv the average velocity during the fourth second v the velocity at any time t vi the initial velocity vii the velocity after 3 seconds viii when and where, if at all, the particle is stationary.Give answers correct to 2 decimal places where appropriate.

7 We 2 The position, x(t) m, of a particle moving in a horizontal straight line from an origin O, at any time t s is given by:

x(t) = (5 − t)(t + 1), 0 ≤ t ≤ 5a Find the initial position.b Find when and where the velocity is 0.c Find the average velocity of the particle during the first 4 seconds.d Sketch a position–time graph for the particle.e Find the average speed in the first 4 seconds by first sketching a position–time line.

8 For each of the following position–time rules, where x cm is the position at any time t seconds, find: i the position after 3 seconds ii when, if at all, the velocity is 0 iii the distance travelled in the first 3 seconds.a x = 2t2 − 8 b x = t3 − 6t2 + 9t + 1 c x = t − 4 loge (t + 1) − 2

d xt= −

12

cosπ

e xt= +

−43

1tan

(Where appropriate, round answers to 2 decimal places.)

9 We 3 For each of the following, find i the velocity and ii the acceleration at any time t.

a x = t3 + 2t2 − 5t b x t t= − +4 2 c x = et + 2e−t

d x = t2 − loge (t + 3) e xt=

−sin 1

3f x = t(5 − t)(t + 2)

10 The height of a projectile h metres above the ground t seconds after it is fired vertically into the air from a tower is given by:

h = 50 + 30t − 5t2

Find:a the average velocity during the first 2 secondsb the time it takes to reach its greatest heightc the greatest heightd the acceleration at any time.

11 The position of an object relative to a reference point O is given by:

x t t t= − +13

3 23 8 , t ∈ [0, 5], where x is in metres and t is in seconds.

a Find the initial position, velocity and acceleration.b Find when and where the object is at rest.



c Find when and where the acceleration is 0.d Sketch i the velocity–time graph and ii the acceleration–time graph.

12 The displacement from O of a particle travelling in a straight line is:

x = t3 − 6t2 + 11t − 6, where x is in cm and t is in seconds.

a At what time does it pass through the origin?b Find the velocity at any time.c When is the velocity 0?d Find the minimum velocity.e Sketch the velocity–time graph for the first 4 seconds.f Find the distance travelled in the first second.

13 An object travelling in a straight line has its displacement from the origin given by:x = 2 cos (3t − 1) + 3

a Find the minimum and maximum displacement.b Find when the velocity is first equal to 0.c How long after it is first at rest is it next at rest?d Find the acceleration in terms of the displacement.

14 Find the velocity and acceleration at any time t for each of the following:

a x = 40 − 5t2 + 2t − t3 b x = 8 sin π t4

c x = 2 + loge (t + 1)

d x = 3e t + 2e–t e x = 5t4 − 2t3 + 4t f x = 2t cos (3t)

g x t e= − −

−

8 10 112 h x

t

t= +1

2

15 We 4 If the position of an object moving in a straight line is given by: x = 2 sin (3t) + 3 cos (3t), find:a the velocity and acceleration at any time tb the initial position, velocity and accelerationc the acceleration in terms of xd the maximum displacement from the origin.

16 A body is projected vertically upward so that its height, h metres, at any time t seconds is given by:

h e t= −

−

−

200 1 4014

Find:a the velocity and acceleration at any time tb the initial height, velocity and accelerationc the maximum height reachedd the acceleration in terms of the velocity.

Using antidifferentiationIf the acceleration, a, of a particle is known in terms of time t, then the velocity, v(t), can be found by antidifferentiation. Thus,

v t a dt c( ) = +∫ 1

where c1 is a constant due to integration; generally it is the initial velocity, v(0).

8bDifferentiate

Displacement(or position) Velocity Acceleration

Antidifferentiate

Differentiate

Antidifferentiate

378

Similarly, the position, x(t), can be found by antidifferentiating velocity with respect to time:

x t v dt c( ) = +∫ 2

where c2 is a constant due to integration; generally it is the initial position, x(0).The constant of antidifferentiation can be determined if initial conditions are given.

v t a dt c( ) 1= +∫x t v dt c( ) 2= +∫

WorKed example 5

The velocity of an object which is initially 3 metres left of O is given by:

v(t) = 3t2 − 2t − 5 m/s

Find:a the displacement from O at any time tb the acceleration at any time tc when the object is at restd the distance travelled in the first seconde the acceleration when the velocity is 0.

ThinK WriTe

a 1 Antidifferentiate with respect to t, to find x. a v(t) = 3t2 − 2t − 5

x(t) = ∫ (3t2 − 2t − 5) dt + c

= t3 − t2 − 5t + c

2 Substitute the initial condition t = 0 and x = −3 into the equation.

When t = 0, x = −3.−3 = 03 − 02 − 5(0) + c

3 Solve the equation for c. −3 = c

4 State the rule for x. x(t) = t3 − t2 − 5t − 3

b Find the acceleration v.(t) by differentiation.

That is, v tddt

v t( ) [ ( )]= .

b v tddt

t t( ) ( )= − −3 2 52

v.(t) = 6t − 2

c 1 Object is at rest when v(t) = 0. c v(t) = 3t2 − 2t − 5 = 0

2 Factorise the LHS. (3t − 5)(t + 1) = 0

3 Solve for t. t = 53 or t = −1, but the domain is

t ≥ 0, so only the first solution is valid.

4 State the answer (t cannot be negative). The object is at rest after 1 23 seconds.

5 Verify using the graph of the velocity function on a CAS calculator.

d 1 The velocity is always negative during the first second and so the distance travelled will equal the magnitude of the displacement during the first second. The object is always moving to the left during the first second.

d At t = 0, v = −5 m/s and at t = 1, v = −4 m/s. Also, v ≠ 0 during the first second. Thus, the distance travelled will equal the magnitude of the displacement during the first second.



2 The displacement is given byx(t) = t3 − t2 − 5t − 3.The distance is thus x(1) − x(0).

x(t) = t3 − t2 − 5t − 3Distance travelled = | x(1) − x(0) | = | (13 − 12 − 5 − 3) − (−3)| = | −5 | = 5 The object moves 5 m to the left during the fi rst second.

e 1 Find when the velocity is 0 by solving v(t) = 0 as in part c .

e v(t) = 3t2 − 2t − 5 = 0(3t − 5)(t + 1) = 0

2 Accept only positive solutions. t = 53 or t = −1, but the domain is

t ≥ 0, so only t = 53 is valid.

3 Substitute the solution into the equation for acceleration.

advdt

=

v(t) = 3t2 − 2t − 5a(t) = 6t − 2

When t a= = × −53

536 2, .

4 Evaluate for a. a = 10 − 2= 8

5 State the answer. When the velocity is 0, the acceleration is 8 m/s2.

WorKed example 6

The acceleration of a particle moving in a straight line is given by:dvdt

e tt= − +5 6 4 cm/s2, where v is the velocity at any time.

If the particle starts at the origin with a velocity of −1 cm/s, fi nd:a the velocity at any time tb the displacement x(t) from the origin at any time tc the displacement from the origin after 1 second.

ThinK WriTe

a 1 Antidifferentiate the acceleration, dvdt

, to fi nd v. advdt

e tt= − +5 6 4

v t e t dt ct( ) ( )= − + +∫ 5 6 4

= 5et − 3t2 + 4t + c

2 Substitute the initial condition v = −1 when t = 0.

When t = 0, v = −1.−1 = 5e0 − 3(0)2 + 4(0) + c

3 Solve for the constant of antidifferentiation. −1 = 5 + cc = −6

4 State the velocity. Therefore, v(t) = 5et − 3t2 + 4t − 6

b 1 Antidifferentiate v with respect to t. b x t e t t dtt( ) ( )= − + −∫ 5 3 4 62

= 5et − t3 + 2t2 − 6t + c

eBookpluseBookplus

Tutorialint-0401

Worked example 6

380

2 Substitute the initial condition x = 0 when t = 0.

When t = 0, x = 0.0 = 5e0 − (0)3 + 2(0)2 − 6(0) + c

3 Solve for the constant of antidifferentiation. 0 = 5 + cc = −5

4 State the displacement. Therefore, x(t) = 5et − t3 + 2t2 − 6t − 5c 1 Substitute t = 1 into the rule for x. c x(1) = 5e1 − (1)3 + 2(1)2 − 6(1) − 5

2 Evaluate. = 5e − 1 + 2 − 6 − 5= 5e − 10

3 Give an approximate answer correct to a useful number of decimal places.

Therefore, the displacement from O after 1 second is approximately 3.59 cm.

4 Verify, using a graph of x(t) on a graphics calculator.

WorKed example 7

The acceleration of a body travelling in a straight line is given by:a(t) = 6t − 2 m/s2; when t = 0, x = 0 and v = −1.

a Find the displacement at any time t. b Find the distance travelled in the fi rst 3 seconds.

ThinK WriTe

a 1 a

2 State the velocity at any time. v(t) = 3t2 − 2t − 1

3


To determine the displacement at any time t, fi rst fi nd the velocity by antidifferentiating a with respect to t, using a CAS calculator by completing the following steps. To do this, on the Main screen, complete the entry line as:Defi ne a(t) = 6t − 2

( ( ))a t( (a t( ( dt∫Press E after each entry.To fi nd the constant of integration, complete the entry line as:Defi ne v(t) = 3t2 − 2t + csolve(v(0) = −1, c)Press E after each entry.

Antidifferentiate v to fi nd the displacement x, completing the entry line as:

( )( )3 2( )( )1( )2( )3 2( )2( )3 2( )( )t t( )( )3 2( )t t( )3 2( )( )3 2( )2( )3 2( )t t( )3 2( )2( )3 2( )dt( )− −( )( )t t( )− −( )t t( )( )3 2( )t t( )3 2( )− −( )3 2( )t t( )3 2( )∫Then press E.The constant of integration can be found by completing the entry lines as:Defi ne x(t) = t3 − t2 − t + dsolve(x(0) = 0, d )Press E after each entry.


4 State the displacement at any time t. x(t) = t3 − t2 − t

b 1 b

2 Write the solution given thatDistance = | x(3) − x(0) | + | x(1) − x(0) |.

Solving v(t) = 0, for t gives

t = 0 as t ≠−13

x(0) = 0x(1) = −1x(3) = 15Distance = | 15 − −1 | + | −1 − 0 | = 17Therefore the body travels 17 m in the fi rst 3 seconds of motion.

v1. (t) = ∫ a dt + c. The instantaneous velocity can be found by antidifferentiating the

instantaneous acceleration. The value of c is found by knowing the velocity at a specifi c time, usually at time t = 0.

x2. (t) = ∫ v dt + c. The displacement and hence the position can be found by

antidifferentiating the instantaneous velocity. The value for c is found by knowing the position at a specifi c time, usually at time t = 0.

remember

Using antidifferentiationA CAS calculator may be used to assist in calculations, or verify answers in the following exercise where appropriate.

1 We 5 The velocity of an object which is initially 2 m right of O is given by:

v(t) = 6t2 − 2t − 8 m/s

Find:a the position x(t) from the origin at any time tb the acceleration x at any time tc when the object is at restd the distance travelled in the fi rst seconde the acceleration when the velocity is 0.

exerCiSe

8b

To fi nd when the object might have changed direction, fi nd t when v = 0.Complete the entry line as:Defi ne x(t) = t3 − t2 − tDefi ne v(t) = 3t2 − 2t − 1solve(v(t) = 0, t)Press E after each entry.

Since t t≠ =t t≠ =t tt t≠ =t t−13

1,≠ =,≠ =t t≠ =t t,t t≠ =t t

Complete the entry line as:x(0)x(1)x(3)Press E after each entry.

382

2 The velocity of an object moving in a straight line, which is initially positioned at x = −1, is given by:

dxdt

t t= − +2 9 42 cm/sFind:a the position at time t b the acceleration at time tc when the object is at rest d the distance travelled during t = 3 to t = 5e the average speed during the time interval t = 3 to t = 5f the average velocity during the time interval t = 3 to t = 5.

3 The acceleration of a body moving in a straight line is given by:

a(t) = 6 − 12t m/s2

Initially, the body is at rest at x = 4. Find:a the velocity at any time b the position at any timec the time and position when the velocity is 0d the distance travelled in the first 3 seconds.

4 We 6 The acceleration of a particle travelling in a straight line is given by:

dvdt

e tt= − +4 2 32

cm/s2

If the particle starts from the origin with a velocity of 2 cm/s, then find:a the velocity at any time tb the displacement from the origin at any time tc the displacement from the origin after 3 seconds (correct to 2 decimal places).

5 mC A particle travelling in a straight line has an initial velocity of −1 m/s and initial displacement of −2 m. The acceleration of the particle is given by:

x = 6t − 2a The particle is stationary when t is equal to:

A 13

s B 2 s C − 13

s D 3 s E 1 s

b The particle is at the origin when t is equal to:A 0 s B 2 s C −1 s D 3 s E 1 s

6 mC If the velocity of a body moving in a straight line is:

x = t2 + 2t − 3 cm/s

then the distance, in cm, travelled in the first 2 seconds is equal to:

A 1 23

B 21

3C 4 D 1 E 3

7 A body is travelling in a straight line with a velocity given by:

dxdt

t

t=

+2 1m/s

If the body starts at the origin, find:a the initial velocityb the position at any timec the acceleration at any timed the maximum velocitye the minimum acceleration.

8 The acceleration of a stone falling from rest from a height of 100 m is given by:

x et= − −

10 m/s2

a Find the velocity at any time t.b Sketch a velocity versus time graph.c Estimate the time taken, to the nearest second, to reach the ground.



9 The velocity of a particle is given by:

dxdt t

=+4

1 2m/s

where the initial displacement from the origin is 1. Find:a the position at any time t b the maximum velocity and when it occursc the acceleration at any time t d the acceleration when t = 1.

10 The acceleration of a particle is:a(t) = 4 sin 2t where t = 0, v = −2 and x = 0

Find:a the velocity at any time t b the position at any time tc the acceleration in terms of the position x d the acceleration in terms of the velocity v.

11 For each acceleration given below, find: i the velocity ii the position at any time t.a x = 6t − 4; when t = 0, x = 1 and x = 0

b x = 3t + 4; when t = 0, x = 6 and x = 0.8

c dvdt

et

= 2 2 ; when t = 0, v = −1 and x = 0

d dvdt

t

t

=−( )

,

4 232

, 0 ≤ t < 2; when t = 0, v = 1 and x = −2

e at

=+

− 3

1 2( ); when t = 0, v = 0 and x = 5

f ae

e

t

t=

+1; when t = 0, v = 0. Find the velocity only.

12 A missile is fired vertically up from the ground with an initial velocity of 20 m/s. The acceleration at any time is given by:

xt= +−102

m/s2

Find:a the height of the missile after 2 secondsb the height of the missile after 4 secondsc the direction the missile is travelling when t = 4.

13 The velocity of a tram travelling on a straight line between two stops is given by:

vt=

1630

sinπ

m/s

Find:a the time lapse between stopsb the distance travelled between stopsc the maximum velocity of the tram and when it occurs.

14 We 7 The acceleration of a body travelling in a straight line is given by:

xt

=+

12

2 1 3( ) cm/s2

and x = 0 cm and x = 0 cm/s when t = 0.a Find the position at any time t.b Find the distance travelled in the fi rst 2 seconds.

u = 20 m/s

384

15 The acceleration of an object starting at rest from the origin is dvdt

= 3 + v, where v is the velocity.

a Find the velocity as a function of time. b Find the position as a function of time.

16 The velocity of a particle starting at the origin is given by dxdt

x= −−

43

2

, where x is the displacement.a Find x as a function of t. b Find the maximum displacement from O.

motion under constant accelerationIf the acceleration is constant, mathematical formulas can be derived for particles travelling in straight lines. Suppose a particle is travelling in a straight line with constant acceleration a. If it is initially at the origin and has an initial velocity of u:

Then dvdt

= a where v is the velocity at any time t

so v a dt c= +∫ v = at + cWhen t = 0, v = u u = a × 0 + c c = uTherefore, v = u + at. [1]

We can further write the displacement in terms of v.

Thus: vdxdt

=

dxdt

u at= +

Thus: x u at dt= +∫ ( )

x ut at c= + +12

21

When t = 0, x = 0, indicating that the object starts at the origin.

Thus: 0 0 012

21= + +u c( ) ( )

c1 = 0

Therefore, x ut at= + 12

2, which is usually written as:

s ut at= + 12

2, where s is the displacement [2]From equation [1]: at = v − u

Transposing to make a the subject gives:

av u

t= −

Substituting into equation [2]:

s ut v utt

= + −12

2

( )

s ut v u t= + −12 ( )

= + −ut vt ut12

12

x u v t= +12 ( ) [3]

Here the distance travelled in a time, t, is equal to the average speed during that time interval,12

( )u v+ , multiplied by the magnitude of the time interval t; this is a familiar result.

8C

O

v = ux = 0t = 0

(or s = ?)v = ?x = ?t = ?

x



By making t the subject from equation [1] we get:

tv u

a= −

Substituting into equation [2]:

s uv u

aa

v u

a= − + −( ) ( )1

2

2

2

suv u v u

a= − + −2 2

2

2 2( )

suv u v uv u

a= − + − +2 2 2

2

2 2 2

2as = v2 − u2

or v2 = u2 + 2as [4]

The four formulas below can be applied to practical problems involving motion in a straight line with constant acceleration a.

1. v = u + at 2. s ut at= + 12

2

3. s u v t= +12

( ) 4. v2 = u2 + 2as

Notes1. Retardation or deceleration implies that the acceleration is negative.2. Remember that acceleration due to gravity is g = −9.8 m/s2 upwards.3. The variable s is the displacement of the object. This is not necessarily the distance

travelled by the object.

WorKed example 8

A particle moves from rest with a constant acceleration, reaching a speed of 20 m/s in 8 seconds. Find:a the acceleration b the distance travelled by the particle.

ThinK WriTe

a 1 List the given information and what has to be found.

a Given: u = 0 v = 20 m/s t = 8Require: a = ?

2 Select an appropriate formula to solve the problem. v = u + at

3 Substitute u = 0, v = 20, and t = 8 into the formula. Substitute u = 0, v = 20 and t = 8.

4 Solve the equation for a. 20 = 0 + 8a

a = 208

= 2.5

5 State the solution. Therefore, the acceleration is 2.5 m/s2.b 1 List the given information and what has to be

found.b Given:

u = 0 v = 20 t = 8Require: s = ?

2 Select an appropriate formula to solve the problem.

s u v t= +12 ( )

eBookpluseBookplus

Tutorialint-0402

Worked example 8

386

3 Substitute u = 0, v = 20 and t = 8 into the formula.

s = +12 0 20 8( )

4 Calculate s. s = 805 State the solution. Therefore, the particle travels 80 metres.

WorKed example 9

A ball is thrown vertically upward from a platform 16 metres above the ground and has an initial velocity of 24.5 m/s. Find:a the maximum height above the platform that the ball reachesb the time taken to reach the maximum heightc the time taken for the ball to fall to the ground from the maximum heightd the total time the ball is in the aire the speed of the ball when it returns to the level of the platformf the speed of the ball when it hits the ground.

ThinK WriTe

a 1 Sketch a diagram of the motion of the ball.Consider up to be positive and down to be negative.

a

16 m

Ground

Platform

a = −9.8 m/s2v = 0

t = 0u = 24.5

2 At its maximum height the velocity of the ball, v, is 0.

v = 0 at the ball’s maximum height.

3 List what is given and what has to be found.

Given: u = 24.5 v = 0 a = −9.8Require: s = ?

4 Select an appropriate formula. v2 = u2 + 2as5

6 Write the solution. Solving 02 = 24.52 + 2 × (−9.8) × s for s gives s = 30.625 m.Therefore, the maximum height reached above the platform is 30.625 metres.


Using a CAS calculator to solve the equation for s, the displacement, complete the entry line as:solve(02 = (24.5)2 + 2 × −9.8 × s, s)Then press E.


b 1 List the given information and what has to found.

b u = 24.5a = −9.8v = 0t = ?

2 Select an appropriate formula. v = u + at

3

4 State the solution. Solving 0 = 24.5 − 9.8 × t for t gives t = 2.5.Therefore, the ball takes 2.5 seconds to reach its maximum height.

c 1 Sketch another diagram for the ball falling from its maximum height.

c

Ground

Platform

Max. heightu = 0

16 m

30.625 m

a = −9.8 m/s2

2 List the given information and what has to be found.Both a and s are negative as their direction is downwards.

Given: u = 0 a = −9.8 s = −(30.625 + 16) = −46.625Require: t = ?

3 Select an appropriate formula. s ut at= + 12

2

4

Using a CAS calculator, solve the equation for t. Complete the entry line as:solve(0 = 24.5 − 9.8 × t, t)Then press E.

Using a CAS calculator, solve the equation for t. Complete the entry line as:solve(−46.625 = 0 × t + 1

2 × −9.8 × t, t)

Then press E.

388

5 Write the solution. Solving −46.625 = 0 × (t) + 12 × (−9.8) × t2 for t gives

t = 3.085.Therefore, it takes approximately 3.085 seconds for the ball to fall to the ground from its maximum height.

d 1 Add the time travelling up to the time travelling down.

d Total time = 2.5 + 3.085= 5.585

2 State the answer. Therefore, the ball is in the air for approximately 5.585 seconds.

e 1 List the given information and what has to be found.

e u = 0a = −9.8s = −30.625v = ?

2 Select an appropriate formula. v2 = u2 + 2as3

4 Write the solution. Solving v2 = 02 + 2 × −9.8 × −30.625 for v gives v = −24.5 or v = 24.5.Since the object is travelling downward, v = −24.5.Therefore, the speed of the ball when it returns to the level of the platform is 24.5 m/s.

f 1 List the information and what has to be found.

f u = 0a = −9.8s = −46.625v = ?

2 Select an appropriate formula. v2 = u2 + 2as

3


Using a CAS calculator to solve the equation for v, complete the entry line as:solve(v2 = 02 + 2 × −9.8 × −30.625, v)Then press E.

Using a CAS calculator to solve for v, complete the entry line as:solve(v2 = 02 + 2 × −9.8 × −46.625, v)Then press E.


4 Write the solution. Solving v2 = 02 + 2 × −9.8 × −46.625 for v gives v = −30.23 or v = 30.23.Therefore, the ball hits the ground with an approximate speed of 30.23 m/s.

When the acceleration a is a constant during an interval of time t, an object will undergo a displacement s. The object has a velocity u at the start of the time interval and a velocity v at the end of the time interval. When motion has a constant acceleration the following formulas can be used:

v1. = u + at

s ut at= + 12

22.

s u v t= +1

2( )3.

v4. 2 = u2 + 2as

remember

motion under constant acceleration 1 Copy the following table, which displays data for objects travelling with constant acceleration.

Complete your copy by using the kinematic equations.

u v a s t

a 0 10 5

b 30 45 2

c −6 8 20

d 18 −1 12

e 4 1.5 4

f 50 −3 24

2 mC A particle moves from rest with constant acceleration of 2 m/s2. Answer the following questions on the particle, which has travelled 25 metres.a The velocity of the particle can be found using the formula:

A v = u + at B s ut at= + 12

2 C s u v t= +12 ( )

D A = πr2 E v2 = u2 + 2as

b The velocity at this time is equal to:A 12.5 m/s B 10 m/s C 25 m/s D 5 m/s E 50 m/s

c The time taken for the particle to reach a speed of 20 m/s is:A 5 s B 40 s C 15 s D 10 s E 2 s

d The distance travelled when it reaches 20 m/s is:A 100 m B 20 m C 1600 m D 5 m E 200 m

3 We8 A particle moving from rest with constant acceleration reaches a speed of 16 m/s in 4 seconds. Find:a the acceleration b the distance travelled.

exerCiSe

8C

390

4 An object travelling at 8 m/s accelerates uniformly over a distance of 20 metres until it reaches a speed of 18 m/s. Find:a the acceleration b the time taken.

5 a A racing car accelerates constantly from rest and covers a distance of 400 metres in 10 seconds. Find its velocity at the end of the 400 metres.

b Another car travels the 400 metres with a constant acceleration of 10 m/s2. Find its time for the 400 metres.

6 A train travelling at a constant speed decelerates uniformly for 30 seconds over a distance of 270 metres, coming to a stop. Find:a the initial speed b the acceleration.

7 A parachutist free-falls from an aircraft for 6 seconds. If the acceleration due to gravity is 9.8 m/s2 downwards, find:a the speed of the parachutist after 6 seconds b the distance travelled after 6 seconds.

8 We9 A ball is thrown up from the ground with an initial velocity of 19.6 m/s. The acceleration due to gravity is 9.8 m/s2 downwards, that is, −9.8 m/s2. Find:a the maximum height attained by the ballb the total time taken for the ball to return to the ground.

9 A stone is dropped from a bridge which is 39.2 metres above a river.a How long does it take the stone to reach the water?b What is its speed on impact?

10 A ball is dropped from a tower and reaches the ground in 4 seconds. Find:a the height of the tower b the velocity of the ball when it hits the ground.

11 A particle is projected vertically up from the top of a building that is 50 metres above the ground. If the initial speed of the particle is 28 m/s, find:a the maximum height, above the ground, that it reachesb total time taken to reach the groundc the speed of the particle when it reaches the ground.

12 A train travels a distance of 1800 metres in 90 seconds while accelerating uniformly from rest. What is its velocity at the end of 500 metres?

13 A car accelerates uniformly from rest, increasing its speed from 5 m/s to 25 m/s in 10 seconds. Find:a the accelerationb the distance travelled, from rest, in 12 secondsc the time taken to increase its speed from 15 m/s to 30 m/s.

14 A sprinter accelerates uniformly to his top speed after running 30 metres of a 100-metre race. He maintains this speed for the remainder of the race and takes 10.4 seconds to complete it. Find:a the top speed of the athleteb the time taken to reach the top speed.

15 A tram is travelling at 16 m/s when the brakes are applied, reducing the speed to 6 m/s in 2 seconds. Assuming the retardation is constant, find:a the accelerationb the distance travelled 2 seconds after the brakes are appliedc how long after applying the brakes the tram comes to a stopd the braking distance of the tram.

16 A car moving from rest with uniform acceleration takes 12 seconds to travel 144 metres. What is its speed after 6 seconds?

u = 0, s = 400 m



17 A tram moving with uniform acceleration has a speed between two points of 4 m/s and 10 m/s respectively. What is the speed half way between the two points? What is the average speed?

18 A bus travels 60 metres in 10 seconds and the next 60 metres in 15 seconds. If the acceleration is constant, find:a how much further it will travel before coming to restb how many more seconds it takes before coming to rest.

19 A juggler throws balls vertically into the air so that they rise to a height of 4.4 metres above the ground. He fails to catch one and it hits the ground with a speed of 1.155 times that of its initial speed. Find:a the speed of projection of the ballb the height from which the ball is thrownc the total time the ball is in the air.

20 An object is projected vertically up from a 14-metre tower and reaches the ground 4 seconds later.a What is the projection speed of the object?b What is the maximum height above the ground that

is attained by the object?

Velocity–time graphsVelocity–time graphs provide visual information about acceleration, velocity, distance and, of course, the time at which events occur:

1. Instantaneous acceleration = dvdt

, the gradient of the v–t graph.

2. Instantaneous velocity = ordinate at a given time.

3. Distance travelled = | |v t dt( )∫ = area bounded by the graph and the t-axis.

Notes1. When measuring distance from a velocity–time graph, the area is measured. When measuring

the displacement, the precise integral is evaluated. Areas below the t-axis are unsigned; that is, they are always positive as distance cannot be negative. However, displacement can be either positive or negative.

2. Area of a rectangle = L × W

3. Area of a triangle = bh2

4. Area of a trapezium = +( )a b h2

WorKed example 10

Use the velocity–time graph to fi nd:a the average acceleration during the fi rst 6 secondsb the average acceleration between t = 20 and t = 24c the total distance travelled during the 24 seconds.

4.4 metres

8dv

t0 t1

v1

v(t) dt = distancet1

A = ∫0

Area (A)

eBookpluseBookplus

Interactivityint-0349

Velocity–timegraphs

v (m/s)

Area2

Area1

t (s)0 6 20 24

8

20

392

ThinK WriTe

a 1 Find the gradient between the points (0, 8) and (6, 20).

a a = −−

=

=

20 86 0

2

126

2 State the acceleration. Therefore, the average acceleration in the first 6 seconds is 2 m/s2.

b 1 Find the gradient between (20, 20) and (24, 0). b a = −−

=

=

−

−

0 2024 20

2045

2 State the acceleration. Therefore, the average acceleration is −5 m/s2, which is a retardation of 5 m/s2.

c 1 Use the formula for the area of a trapezium to calculate the area of the two trapezia.

c Aa b h= +( )

2

2 Calculate the area of the trapezium from t = 0 to t = 6.

Area18 20 6

228 6

284

= +

= ×

=

( )( )

3 Calculate the area of the trapezium from t = 6 to t = 24.

Area214 18 20

232 20

2320

= +

= ×

=

( ) ( )

4 Add the two areas. Total area = 84 + 320 = 404

5 State the total distance travelled. Therefore, the total distance travelled is 404 metres.

WorKed example 11

Sketch a velocity–time graph for the motion described as follows:‘A particle accelerates uniformly from rest to a speed of 12 m/s in 10 seconds. It maintains this speed for another 10 seconds then decelerates at a uniformly decreasing rate to rest in a further 6 seconds.’

ThinK WriTe

1 The graph starts from (0, 0) as a straight line to the point (10, 12) (since the gradient is constant).

v (m/s)

t (s)0 10 20 26

122 The graph continues from (10, 12) to (20, 12) as a

horizontal line (since the gradient is 0).

3 The graph continuing from (20, 12) to (26, 0) has to be part of a parabola because the rate of change of deceleration is uniform.



4 Furthermore, it has to be a positively shaped parabola because the deceleration is decreasing (that is, the gradient must be getting less steep).

5 Sketch the graph.

WorKed example 12

A train which is initially at rest at one station accelerates uniformly to a speed of 75.6 km/h. It maintains this speed for 1 minute, then decelerates at twice the magnitude of the acceleration to rest at the next station. The two stations are 1535 metres apart.a Sketch a velocity–time graph showing the motion of the train between the two stations.b How long does it take the train to travel between the two stations?

ThinK WriTe

a 1 Convert 75.6 km/h to m/s by dividing by 3.6.

a 75.6 km/h = 75 63 6

..

m/s

= 21 m/s

2 Let the time taken for the initial acceleration be T.

Let the acceleration time = T.

3 Find the time for deceleration by dividing T by 2, since the deceleration is twice the magnitude of the acceleration.

Deceleration time = T2

(given that the

deceleration is twice the magnitude of the acceleration).

4 Sketch the velocity–time graph. v (m/s)

1260

60Tt(s)0 T T + 60

21

T + 60 +

–2T

= + 60 –23T–

2T

b 1 Use the velocity–time graph to fi nd the distance travelled at constant speed by multiplying 60 by 21 (the area of the rectangle).

b Distance travelled at constant speed = 21 × 60 = 1260 m

2 Find the total of accelerating distances by subtracting 1260 from total distance.

Total acceleration distance = 1535 − 1260 = 275 m

3 Express the total of the accelerating distances as the sum of the area of the triangles before and after the rectangular area.

or total accelerating distance

= +12

12

21 212

( ) ( )TT

4 Simplify this expression.

= +

=

212

214

634

T T

T

eBookpluseBookplus

Tutorialint-0403

Worked example 12

394

5 Equate the two areas. So 63

4275

T =

6 Solve the equation for T. 63T = 1100 T = 17.46

7 The total time is the sum of 60, T and T2

. Total time = 60 + T + T2

8 Substitute T = 17.46 into the expression for total time.

= 60 + 17.46 + 17 462.

9 Evaluate the total time. = 60 + 26.19 = 86.19

10 State the solution. Therefore, the train takes 86.19 seconds or 1 min 26.19 s to travel between the two stations.

WorKed example 13

A lift operates between two floors within a building. The two floors are 80 metres apart. The lift can travel with a constant acceleration of 2 m/s2, decelerate at 3 m/s2 or travel at constant speed. If the minimum travelling time between the two floors is 15 seconds, find the maximum speed of the lift.

ThinK WriTe

1 Let the maximum speed be v1. Let maximum speed = v1.

2 List the given information and what has to be found.

Given: a = 2 u = 0 v = v1

Require: t, the time to obtain a speed v1.

3 Select an appropriate formula. Use v = u + at.

4 Substitute u = 0, a = 2, v = v1 into the formula.

Substitute u = 0, v = v1, a = 2. v1 = 0 + 2t

5 Make t the subject. tv

= 1

26 List what is given for the deceleration phase

and use v = u + at again to find an expression for the time taken for the lift to decelerate.

Substitute u = v1, v = 0, a = −3 (deceleration phase) into v = u + at.

7 Substitute u = v1, a = −3, v = 0 into the formula. 0 = v1 − 3t

8 Make t the subject. 3t = v1

tv= 1

39 Sketch a velocity–time graph, showing the

three phases of the lift’s motion.

––3

v (m/s)

t(s)0 15

v1

––2v1 v1



10 Find an expression in terms of v1 for the time the lift travels at constant speed.

Time travelled at constant speed

= − −

= −

=−

152 3

1556

90 56

1 1

1

1

v v

v

v

11 Calculate the area of the trapezium in terms of v1, and equate to 80, the distance travelled.

Total distance travelled:

80

1590 5

6

2

11

=+ −

vv

, (area of trapezium)

12

13 Write the solution. v1 = 29.49 or 6.51

14

15 Write the solution. When vv

1129 49

90 56

9 575= − = −. , . .

When vv

116 51

90 56

9 575= − =. , . .

As time must be positive, v1 = 6.51.Therefore the maximum speed of the lift is approximately 6.51 m/s.

Using a CAS calculator, solve for v1, by completing the entry line as:

solve 15

90 56

280

+ −

× =

v

v v80v v80× =v v× = ,v v,v v

Then press E.

The time during which the lift moves with constant speed must be positive. Substitute

each value for v1 into 90 56

1− v , by completing the entry line as:90 5

629 491− =v

v| .

Then press E.Repeat for v1 = 6.51

396

Instantaneous acceleration 1. = the gradient of the v–t graph.

Velocity at time 2. t = the ordinate at time t.

Distance travelled in time interval 3. δ t = the area bounded by the graph, the t-axis and the time interval chosen δ t. Note that area is always a positive number.

Displacement during time interval 4. t1 to t v t dt x t x t2 2 1= = −∫ ( ) ( ) ( ).

remember

Velocity–time graphsA graphics or CAS calculator may be used to assist in calculations or to verify answers in the following exercise where appropriate.

1 We 10 Use each of the following velocity–time graphs below to find: i the average acceleration during the first 10 seconds ii the distance travelled in the first 30 seconds.

a v

t(s)0 10 30

12

(m/s) b v (m/s)

t(s)0 10 30

1520

c v (m/s)

t(s)0 15 30

10

18

d v (m/s)

t(s)0 10 30

1620

e v (m/s)

t(s)0 15 30

10

−10

f v (m/s)

t(s)0 10 20 30

8

2 We 11 Sketch a velocity–time graph for each motion described below.a A particle travels at a constant speed of 14 m/s.b A particle moves from rest with a constant acceleration, attaining a speed of 12 m/s in

15 seconds. It then maintains this speed.c A body accelerates uniformly from rest to a speed of 15 m/s in 10 seconds. It then

decelerates constantly to rest after a further 12 seconds.d An object is initially travelling at 18 m/s. It maintains this speed for 20 seconds, then

decelerates uniformly to a speed of 8 m/s in 10 seconds, thereafter maintaining this speed.

exerCiSe

8d



e From rest, a particle accelerates at a uniformly decreasing rate, reaching a speed of 15 m/s in 15 seconds. It maintains this speed for 20 seconds, then decelerates uniformly to rest in a further 8 seconds.

f A particle accelerates from rest at a constantly increasing rate, reaching a speed of 19 m/s in 11 seconds. After maintaining this speed for 10 seconds, it decelerates at a uniformly decreasing rate, coming to rest after a further 7 seconds.

3 The following velocity–time graphs are for various objects that move in a straight line and start at the origin. In each case find, as a function of time, t: i the velocity ii the acceleration iii the displacementof the objects described by the figures in a–f.a v

t0 10

8

b v

t0 10

14

c

Given that v = at2 + b,where a and b are constants.

(4, 22)v

t0

−2

d

Given that v = at2 + bt + c.

(16, 26)(4, 30)

v

t0

10

e

This is of the form v = a sin (bt) + c.

v

t0

1

3 —4π

– 4( , 4)

( , −2)

π f

This is of the form v = aebt.

(loge 9, 24)

v

t0

8

4 mC The velocity–time graph at right shows the motion of a cyclist travelling on a straight stretch of road.a The initial acceleration is:

A 0.4 m/s2 B 2.5 m/s2 C 2 m/s2

D −2 m/s2 E −0.4 m/s2

b The acceleration from t = 25 to t = 28 is:

A 2.5 m/s2 B − 13 m/s2 C −2 m/s2

D 3 m/s2 E −2.5 m/s2

c The total distance travelled in 35 seconds is:A 429 m B 500 m C 627 mD 408 m E 510 m

d The cyclist travels half of the total distance after:A 12 s B 18 s C 17.3 sD 16.5 s E 15.8 s

v (m/s)

t (s)0 6 25 28 35

9

15

398

5 mC A car starting from rest accelerates uniformly on a straight road to 90 km/h. It maintains this speed for a while, then comes to rest with a uniform retardation of 1.25 m/s2. The total distance travelled is 1 km in a total time of 55 seconds. Use a velocity–time graph to help with the following.a The time taken for the retardation phase is:

A 10 s B 20 s C 72 s D 40 s E 36 sb The time into the journey at which the car starts to decelerate is:

A 35 s B 20 s C 40 s D 10 s E 50 sc The time taken during the acceleration phase is:

A 35 s B 20 s C 25 s D 10 s E 60 sd The distance the car travels at 90 km/h is:

A 500 m B 700 m C 625 m D 600 m E 250 m

6 We 12 A particle starting from rest and accelerating uniformly reaches a speed of 8 m/s in 2 seconds. It maintains this speed for a further 6 seconds and then decelerates uniformly to rest in 3 seconds. Use a velocity–time graph to find the total distance travelled.

7 Two bus stops are 450 metres apart. A bus departs from one with constant acceleration until it reaches a point where it decelerates uniformly at twice the magnitude of its acceleration. This brings the bus to a halt at the other bus stop. The total time taken between the stops is 1 minute.a Sketch a velocity–time graph for the bus.b Find: i the acceleration and ii the maximum speed reached.

8 A bus accelerates and decelerates at 2 m/s2. Its maximum speed is 72 km/h. Find the shortest time over which the bus can travel between two stops 800 metres apart.

9 A train leaves a station from rest with a constant acceleration of 2 m/s2 until it reaches a maximum speed of 18 m/s. It continues at this speed until it is brought to rest under a retardation of 3 m/s2. The distance travelled by the train to the second station is 1485 metres. Using a velocity–time graph, find the time taken to travel between the two stations.

10 We 13 A lift operates over a vertical distance of 120 metres. Its acceleration is 0.75 m/s2 and its deceleration is 1 m/s2 and in between it travels at a constant speed for 13 seconds. The minimum time for the journey is 27 seconds. What is the maximum speed of the lift?



11 A lift operates between the ground floor and top floor of a building, a vertical distance of 180 metres. It can move with a constant acceleration of 1.2 m/s2 or deceleration of 1.6 m/s2 or at constant speed. If its greatest speed is 7.2 m/s, find the least time taken to travel between the ground and top floors.

12 Two cars start from rest at the same place and same time and move in the same direction with constant but different accelerations. If they are 90 metres apart after 60 seconds, find the difference between their speeds at this point.

13 Go-carts A and B are stationary on a 400-metre straight racing track. At t = 0, cart A moves off with a constant acceleration of 2 m/s2 for 10 seconds, after which it moves at constant speed. Cart B sets off at the same time with constant acceleration of 1.5 m/s2 for 16 seconds, then maintains a constant speed. Which cart completes the 400 metres first and how far ahead of the second cart is it when it does so?

14 A stationary police car is passed by a motorcycle travelling at a constant speed of 72 km/h. The police car sets off in pursuit of the motorcycle 5 seconds after the motorcycle has passed it. The police car moves with constant acceleration for a distance of 200 metres, reaching a speed of 90 km/h which it then maintains. How long after the motorcycle first passes the police car does the police car catch up to the motorcycle?

15 A car slows down with a constant retardation from 24 m/s to 16 m/s over a distance of 15 metres. What further distance will it travel before coming to rest?

16 Two cars, A and B, are moving with constant acceleration in the same direction on parallel lanes of a freeway. At t = 0 s, car A passes car B when the speeds are 54 km/h and 36 km/h respectively. Two minutes later car B, travelling at 90 km/h, passes car A.

Find:a the distance travelled by the cars between the times at which they pass each otherb the time after their fi rst passing when they are travelling at the same speed, and this speedc the distance between them when their speeds are the same.

17 A drag-racing car accelerates from rest at 5 m/s2 until it reaches its maximum speed, which it maintains for 2 seconds. It then releases its parachute, which decelerates it at 8 m/s2 until it is brought to rest after travelling 506.25 metres in total. Find:a the maximum speed reached by the carb the braking distance (distance travelled after the parachute is released)c the average speed.

18 Two trains pass one another, travelling in opposite directions on parallel tracks. When the fronts of the trains are in line they are travelling at 12 m/s and 16 m/s and accelerating at 0.5 m/s2 and 1 m/s2 respectively. The length of each train is 136 metres. How long does it take the rear of the two trains to pass?

19 A car travelling at 90 km/h along a highway passes a stationary motor cycle policeman. The policeman starts off in pursuit 5 seconds later, moving for 250 metres with constant acceleration until he reaches a speed of 120 km/h which he maintains.a Find the time taken for the policeman to catch up to the car.As soon as the policeman draws level with the car he indicates to the driver to stop. Both uniformly apply the brakes, and stop 3 seconds later.b How far does the policeman have to walk back from his cycle to the stationary car?

20 Two cars A and B, each moving with constant acceleration, are travelling in the same direction along the parallel lanes of a divided road. When A passes B, the speeds are 64 and 48 km/h respectively. Three minutes later B passes A, traveling at 96 km/h. Finda the distance travelled by A and B at this instant (since they fi rst passed) and the speed of A,b the instant at which both are moving with the same speed, and the distance between them

at this time.

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21 A motorist is travelling at a constant speed of 120 km/h when she passes a stationary police car. She continues at that speed for another 15 seconds before uniformly decelerating to 100 km/h in five seconds and then continues with constant velocity.

The police car takes off after the motorist the instant it passes. It accelerates uniformly for 25 seconds 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 she traveled in that time?

applying differential equations to rectilinear motionIn the previous sections of this chapter the position, velocity and acceleration of bodies moving in straight lines were all represented as functions of time. We will now consider situations where the velocity or acceleration of the body is, for example, a function of its position. This requires the use of various differential equations, depending on the information given and the information required.

Velocity given as a rate of change of positionIf the velocity is given as a function of position and the position is required as a function of time, we proceed as follows: v = f (x)

dxdt

f x vdxdt

= =( ), since

or dtdx f x

= 1( )

, inverting

⇒ = ∫tf x

dx1( )

Remember that:

1. tf x

dx= ∫ 1( )

2. After the antiderivative is found, the equation may then be rearranged to make x the subject.

WorKed example 14

A particle moves in a straight line such that its velocity, v cm/s, is

v xx

( ) =+1

2 cm/s, x > −2. When t = 0, x = 0.

Find:a the position, x, as a function of time, tb the position after 2.5 secondsc the distance travelled in the first 2.5 seconds.

ThinK WriTe

a 1 Express v as dxdt

. a vx

x=+

> −12

2,

dxdt x

=+1

2

2 Invert both sides.dtdx

x= + 2

8e



3 Express t in integral notation. t x dx= +∫ ( )2

4 Antidifferentiate the integrand with respect to x. t x x c= + +12

2 2

5 Substitute t = 0 and x = 0. When t = 0, x = 00 = 0 + 0 + c

6 Solve the equation for the integration constant c. c = 0

7 State the equation for t as a function of x. So t x x= +12

2 2

8 Multiply both sides by 2. or 2t = x2 + 4x

9 Express the RHS as a perfect square to get x by itself, so that x can be expressed as a function of t.

= (x + 2)2 − 4

10 Add 4 to both sides. or (x + 2)2 = 2t + 4

11 Take the square root of both sides. x t+ = ± +2 2 4

12 Subtract 2 from both sides and keep the positive square root only as x > −2 (given.)

x t= + +− 2 2 4, as x > −2

b 1 Substitute t = 2.5. b When t = 2.5,x = + +− 2 2 2 5 4( . )

2 Evaluate x. = +− 2 9= −2 + 3

x = 1

3 State the answer. Therefore, the position after 2.5 seconds is 1, or 1 cm right of O.

c 1 Since vx

=+1

2 and x > −2, the velocity is

always positive.

c Since vx

=+1

2 and x > −2,

v > 0 at all times.

2 Since the velocity is positive the object is moving to the right only and does not turn.

Therefore, the object is travelling to the right (or in one direction) at all times.

3 The distance travelled is the position at t = 2.5 minus the position at t = 0.

Distance = x(2.5) − x(0)

4 Calculate the distance. = 1 − (−2 + 4)= 1 − (−2 + 2)= 1 − 0= 1

5 State the answer. Therefore, the object travels 1 cm in the first 2.5 seconds.

If the velocity is given as a function of position x, and the acceleration is required as a function of x, we proceed as follows. a

dvdt

=

= ×dvdx

dxdt

, using the chain rule.

402

But dx

dt is the velocity, v.

Therefore a vdvdx

=

WorKed example 15

The velocity of a particle moving in a straight line is given by v(x) = 2 + loge (x), x > 0, where x is the position at any time, t. What is the acceleration in terms of x?

ThinK WriTe

1 Write down the velocity. v = 2 + loge (x)

2 Find dvdx

.dvdx x

= 1

3 Multiply v by dvdx

to fi nd the acceleration. a vdvdx

=

4 Simplify the RHS. a x xe= +( log ( ))2 1

ax

xe=

+2 log ( )

acceleration given as a rate of change of velocityIf the acceleration is given as a function of velocity then the position, velocity and acceleration can be found in terms of time, t.That is, if a = g(v), since a

dvdt

=

then dtdv g v

= 1( )

tg v

dv= ∫ 1( )

Remember that:

1. tg v

dv= ∫ 1( )

2. The antiderivative can be rearranged to make v the subject, giving v as a function of time. The position and acceleration can be found by antidifferentiating and differentiating the velocity respectively, as seen earlier.

WorKed example 16

The acceleration of a particle travelling in a straight line is given by dvdt

v= + 1,

where v is the velocity at any time t. At time t = 0, the body is at rest at the origin. Find, as a function of time:a the velocityb the accelerationc the position.

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Worked example 16



ThinK WriTe

a 1 Write the acceleration as given. advdt

v= +1

2 Invert both sides of the differential equation.

or dtdv v

=+1

1

3 Express t in integral form. tv

dv=+∫ 1

1

4

5 Write the solution. tv

dv=+

=∫ 11

loge (| v + 1 |) + c

Solving 0 = loge (| 0 + 1 |) + c for c gives c = 0.So, t = loge (| v + 1 |)Then, solving t = loge (| v + 1 |) for v gives v = −et − 1 or v = et − 1.Since v = 0 when t = 0Then v = et − 1

b 1 Differentiate v with respect to t to fi nd the acceleration (or we could substitute v into the original equation for acceleration).

b advdt

=

ad e

dt

t

= −( )1

2 State the answer. a = et

c 1 Express x as the antiderivative of v. c x e dtt= −∫ ( )1

2 Antidifferentiate the integrand. x e t ct= − +

3 Substitute t = 0 and x = 0, the initial condition given.

When t = 0, x = 00 = e0 − 0 + c

4 Solve for c. c = −1

5 State the answer. x = et − t − 1

Antidifferentiate the integrand using the CAS calculator.Complete the entry line as:

11v

dv+∫

Then press E.Given that t = 0 when v = 0, to fi nd the constant of integration. Complete the entry line as:solve(0 = ln(| 0 + 1 |) + c, c)Then press E.Rearrange the equation to make v the subject by completing the entry line as:solve(t = ln(| v + 1 |), v)Then press E.

404

acceleration given as a function of positionThe following relation is used if the acceleration is given as a function of position, x, and the velocity as a function of x is required:

vddv

v=

12

2

Now a vdvdx

= (seen earlier).

Substituting ddv

v12

2

for v, gives:

addv

v dvdx

=

×1

22

addx

v= 12

2

WorKed example 17

The acceleration of an object travelling in a straight line is given bya(x) = ex + 1 cm/s2, x ≥ 0, where x is the position at any time, t.

The velocity is 1 when x = 0.a Explain why the velocity is always positive.b Find the velocity v(x) in terms of the position x.

ThinK WriTe

a 1 The acceleration, a = ex + 1, is > 0 for all values of x since ex > 0 for all x.

a a = ex + 1 > 0 for all x(since ex > 0 for all x)

2 The velocity is always increasing as x increases since a > 0.

The velocity, v, is always increasing as x increases.

3 The velocity is initially equal to 1 where x = 0. At x = 0, v = 1.

4 The velocity must be greater than or equal to 1. As x increases, v > 1.

5 The velocity is always positive. Therefore, the velocity is always positive.

b 1 Use the rule addx

= 12

2v

to set up a

differential equation for acceleration.

b ddx

12

2v

= ex + 1

2 Express 12 v2 in integral notation. 1

2 v2 = ( )e dxx +∫ 1

3 Antidifferentiate the integrand. = ex + x + c

4 Substitute x = 0 and v = 1 into the equation. When x = 0, v = 1.

12 (1)2 = e0 + 0 + c

5 Solve the equation for c. 12 = 1 + c

c = −1

2

6 Rewrite the rule for 12 v2. So 1

2 v2 = ex + x − 1

2

7 Multiply both sides by 2. or v2 = 2ex + 2x − 1

8 Take the square root of both sides. v e xx= ± + −2 2 1



9 Dismiss the negative square root, as v > 0, and state the solution.

but v > 1

v e xx= + −2 2 1

Summary of acceleration typesThe different forms of acceleration are:

ad x

dt

dvdt

vdvdx

d v

dx= = = =

2

2

12

2

The acceleration equation given will determine which form to use. The following summary may help you decide.

Acceleration a = f(t) a = f(x) a = f(v)

Form of acceleration to use

dvdt

d x

dtor

2

2d v

dx

12

2

dvdt

if the initial

conditions are in terms of t and v

or

vdvdx

if the initial

conditions are in terms of x and v.

If 1. v = f (x), then tf x

dx= ∫ 1( )

.

a vdvdx

=2. and thus: addx

= (12 v2).

If 3. a = g(v), then tg v

dv= 1( )

.

remember

applying differential equations to rectilinear motionA CAS calculator may be used to assist in calculations, or verify answers in the following exercise where appropriate.

1 We 14 A particle moves in a straight line such that its velocity, v cm/s, is:

v xx

( ) ;=−1

2 6 x > 3; x = 4 when t = 0.

Find:a the position, x, at any time, tb the position after 15 secondsc the distance travelled in the first 15 seconds.

2 An object initially at the origin is moving in a straight line and has a velocity given by:v(x) = 2x + 1 cm/s

Find:a the position at any time, t b the velocity at any time, tc the initial acceleration.

exerCiSe

8e

3 The velocity of a body moving in a straight line is given by:

v xx

xe

( )log ( )

,=2

x ≥ e; x = e when t = 0.Find:a x in terms of t b v in terms of tc a in terms of t.

4 We 15 The velocity, v m/s, of a particle moving in a straight line is given by:

v(x) = (x − 1)2, x > 1

Find the acceleration when the velocity is 1 m/s.

5 mC A particle moves in a straight line so that the velocity is given by v = 1 + e−x m/s, and x = 0 when t = 0.a The acceleration in terms of x is:

A 12 x − e−x − 1

4 e−2x B x − e−x C e−2x − e−x

D −e−x − e−2x E x + e−x

b The position, x m, in terms of t is:A 2et − 1 B loge (et + 1)

C loge (2et − 1) D 2t − 1

E e−x + tc The initial acceleration in m/s2 is:

A −1 B − 54

C 0 D 1 E −2

6 An object moving in a straight line has a velocity given by:

v x x( ) = −4 1 2

Find:a the acceleration in terms of x b the position in terms of t if x = 1 when t = 0.

7 The velocity of a body moving in a straight line is given by v(x) = 4 + x2 cm/s, where at time t = 0, x = 2.

Find:a the acceleration in terms of position x b the position in terms of t

c the velocity when t = π24

.

8 A body moves in a line so that its acceleration is (x + 1) m/s2 and when x = 2, v = 3.a Find where the body is when the velocity is 0.b Explain why v is increasing if x > −1. c Find v in terms of x.d Find the body’s velocity when it is at the origin.

9 The acceleration of an object moving in a straight line is 4x cm/s2 and when x = 0 the velocity

is 1. What is the velocity when x = 2?

10 We 16 The acceleration of a particle travelling in a straight line is given by:

dvdt

v= +2

Initially, the object is at rest at the origin. Find:a the velocity in terms of t b the acceleration in terms of tc the position at any time t.

11 The acceleration of an object moving in a straight line is given by:dvdt

v= −4 and v = 3 when t = 2.Find:a v at any time t b the acceleration when t = 3.

exam Tip A common mistake occurs when students simply differentiate the given expressionwith respect to x, confusing dv

dx with acceleration

instead of using a vdv

dt

dx

dt

dv

dx

dv

dx= = × = × .

406 maths Quest 12 Specialist mathematics for the Casio Classpad


12 A body moving in a straight line has an acceleration given by:

av= −1560

2

Initially the body is at rest at the origin.a Find the velocity in terms of t. b Find the acceleration in terms of t.c Find the displacement in terms of t.

13 We 17 The acceleration of an object travelling in a straight line is given by:

ax

=−1

2 2( ) cm/s2, 0 ≤ x < 2, where x is the position at any time.

The velocity is 0 when x = 0.a Explain why the velocity is always positive.b Find the velocity v(x) in terms of the position x.c Find where the particle has a velocity of 3 cm/s.

14 A body at rest starts moving with an initial acceleration of 5 m/s2. The acceleration decreases uniformly with the distance travelled reaching a value of zero when the body has travelled 80 metres. Find the maximum speed of the body.

15 An object is dropped from rest so that the acceleration due to air resistance is 0.2v where v is the speed of the object. The acceleration due to gravity is g m/s2.

a Show that dvdt

= g − 0.2v.

b Find v in terms of t.c Find lim ( )

tv t

→∞, that is, the maximum velocity.

d Find the distance fallen after 5 seconds (to the nearest metre).

16 A cork is projected vertically upwards (take ‘upwards’ to be positive) with an initial speed u m/s against an air resistance kv2, where v is the velocity in m/s at any instant and k is a constant. If x is the height (in metres) of the cork at any time show that:a a = −(g + kv2)and hence fi nd:b the velocity in terms of xc the maximum height reachedd the maximum height if k = 0.002 and u = 30.

17 A particle has a displacement, x = 49t − 225 + 245e−0.2t metres above the ground after a time t seconds. Find:a its initial height above the groundb its velocity v at time t and show that its initial velocity is zeroc its acceleration a at time t, and show that a = 9.8 − 0.2v.d Deduce that the limiting speed is given by 49 m/s.

18 A particle moves back and forth along a straight line track so that its displacement at a time t seconds is given by x = 3 sin (2t) + 4 cos (2t).a Find its initial displacement and initial velocity.b Find its acceleration after π seconds.c If its velocity is v, show that v2 = 4 (25 − x2).d If its acceleration is a, show that a = −4x.

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The falling ball bearing

408

SUmmary

Differentiation and displacement, velocity and acceleration

Position gives the location of a particle relative to a reference point (usually the origin). The variable used is • x.Displacement is change in position • δ x or s. Displacement and position have the same value if a body is initially at the origin and so the two terms are often used to mean the same thing. Hence, the same variable is used, namely x.

Instantaneous velocity, • v or dxdt

or x, is the time rate of change of position or the rate of displacement.

Average velocity (during time interval • δ t) = = =δδ δ

xt

st

displacementtime

.

Speeddistance travelled

time taken=• .

Instantaneous acceleration, • a or x or d x

dt

2

2 or

dvdt

is the time rate of change of velocity.

Using antidifferentiation

v t a dt c( ) = +∫• . The instantaneous velocity can be found by antidifferentiating the instantaneous

acceleration. The value for c is found by knowing the velocity at a specific time, usually at time t = 0.

x t v dt c( ) = +∫• . The displacement and hence the position can be found by antidifferentiating the

instantaneous velocity. The value for c is found by knowing the position at a specific time, usually at time t = 0.

Motion under constant acceleration

When the acceleration • a is a constant during an interval of time, t, an object will undergo a displacement, s. The object has a velocity, u, at the start of the time interval and a velocity, v, at the end of the time interval. When motion has a constant acceleration the following formulas can be used:

v1. = u + at 2. s ut at= + 12

2

s u v t= +12 ( )3. 4. v2 = u2 + 2as

Velocity–time graphs

Instantaneous acceleration • = the gradient of the v–t graph.Velocity at time • t = the ordinate at time t.Distance travelled in time interval • δ t = the area bounded by the graph, the t-axis and the time interval chosen δ t. Note that area is always a positive number.

Displacement during time interval • t1 to t v t dt x t x t2 2 1= = −∫ ( ) ( ) ( ).

Applying differential equations to rectilinear motion

If • v = f (x), then tf x

dx= ∫ 1( )

.

a vdvdx

=• and thus: addx

v=

12

2 .

If • a = g(v), then tg v

dv= ∫ 1( )

.

If • a = h(x), then vdvdx

h x= ( ).



ChapTer reVieW

ShorT anSWer

1 A particle moves in a straight line so that its position relative to a point O, at any time t seconds, is given by x = t2 − 6t + 8, t ≥ 0, where x is in metres.Find:a the time at which the particle’s velocity is zerob the acceleration of the particle when its

velocity is zeroc the distance travelled by the particle in the first

4 secondsd the time and the particle’s distance when its

velocity is −2 m/s.

2 The acceleration, a m/s2, of a body moving in a straight line at any time t seconds (t ≥ 0) is given by the rule a = 8 − 6t. The initial position of the body is 5 m to the right of a point O, and its initial velocity is 6 m/s. Find the particle’s position and velocity after 3 seconds.

3 A car, starting from rest, accelerates from a set of traffic lights at a constant rate for a certain time. The brakes are then applied. The car then decelerates at a constant rate, but at twice the magnitude of its acceleration, before coming to rest at a second set of traffic lights. If the traffic lights are 400 metres apart and the time taken to travel between them is 40 seconds, find:

a the maximum speed reached by the car in km/hb the time at which the brakes are appliedc the acceleration of the car from the first set of

traffic lights.

4 The velocity–time graph above right shows the journey of a person who jumps from a platform into a pool of water.

0

−2

10

1 2

Water

(1, 10)

(4, −2)

Air

3 4 t (s)

v (m/s)

Find:a the height of the platformb the acceleration of the person through the waterc the maximum depth the person reaches in the

waterd the depth the person reaches in the water after

4 seconds.

5 A stone is thrown vertically upwards from ground level with an initial velocity of 29.4 m/s. Calculate:a the velocity and direction of the ball after

4 secondsb the total distance, in terms of g (acceleration

due to gravity), travelled by the stone after it returns to the ground

c the velocity of the stone when it strikes the ground.

6 A cyclist starting from rest travels a total distance of 1800 metres before finishing at rest. During the journey, the cyclist’s velocity increases uniformly until he reaches 16 m/s. This velocity is maintained for 100 seconds and then uniformly decreases to zero.a Sketch the velocity–time graph of the cyclist.b Find the total time of the cyclist’s ride.

410

7 A particle moves in a straight line so that its position relative to a point O, at any time t seconds, is given by x = 2 loge (t + 1), t ≥ 0 where x is in metres. Find the particle’s velocity and acceleration after 2 seconds.

8 The acceleration of a falling parachutist x metres after the parachute is released is given by the rule

a v= −0 132. m/s2, where v is the velocity of the

parachutist at any given time. If the parachutist is falling at 25 m/s when the parachute is released, how much further does he fall before his velocity is reduced to 4 m/s?

9 A lawn bowl is rolled across a level green in a straight line. It is released with a velocity of 5 m/s and the velocity decreases at a rate 0.5 m/s2 due to friction and air resistance. How far does the lawn bowl travel?

u = 5 m/s

10 A bullet fired from a revolver with a muzzle velocity of 70 m/s is subject to a retardation of −0.4x, where x is the distance in metres travelled by the bullet. Find the velocity of the bullet after it has travelled 100 metres.

mUlTiple ChoiCe

Questions 1 to 3 refer to the graph below, which shows the position of a particle moving in a straight line, x, as a function of t. Displacement in the direction ‘right’ is taken to be positive.

(3, 10)

(10, −4)

x (cm)

t (s)0

5

10

8 13

1 The initial position is at:A 8 B 0 C 13 D 5 E −8

2 The particle is travelling to the left when:A 3 < t < 10 B 0 < t < 3 C 8 < t < 10D 10 < t < 13 E t > 10

3 The average velocity between t = 3 and t = 13, in cm/s, is:

A 1.8 B −1 C −1.8 D 1 E 2313

Questions 4 to 6 refer to a particle moving in a straight line, which has a displacement, x cm, from the origin at any time, t seconds, given by:

x(t) = t2 − 6t + 5

4 The initial velocity in cm/s is:A 4 B 0 C −2 D 5 E −6

5 The distance travelled in the first 4 seconds, in cm, is:A 8 B 9 C 10 D 3 E 7

6 The number of times the particle passes through the origin is:A 1 B 2 C 0 D 4 E 3

7 A particle moves in a straight line so that its position at any time is given by:

x = t3 − 2tThe acceleration of the particle when t = 1 is:A 1 B 4 C 0 D 6 E 8

8 The velocity of a body moving in a straight line is v = 3t2 + 5. Initially the body is at −2. The position at time t = 2 is:A 16 B 17 C −4 D 15 E 10

9 A body starts from the origin, travelling in a straight line with a velocity given by:

xt

t t= +

+ +1

2 32

The position of the body at time t = 3 is closest to:A 0.90 B 3.21 C 9 D −2.5 E 0

10 An object is dropped from the top of a 200-metre high building. If the acceleration due to gravity is 9.8 m/s2, what will be the height of the object after 5 seconds?A 151 m B 49 m C 122.5 mD 20 m E 77.5 m

11 A particle initially moving at 6 m/s is subject to a constant retardation of 2 m/s2. The distance, in metres, travelled before coming to rest is:A 27 B 8 C 9D 10 E 12

12 A tram travels 500 metres in 25 seconds when accelerating uniformly from rest. The acceleration, in m/s2, is:A 0.4 B 1.6 C 1.25D −1.2 E 0.625



Questions 13 to 15 refer to the following information. A car accelerates uniformly from rest, reaching a speed of 20 m/s after 5 seconds. It maintains this speed for 13 seconds before decelerating uniformly to rest in 4 seconds.

13 The velocity–time graph representing the motion of the car is:A v (m/s)

t(s)02218

20

B v (m/s)

t(s)022135

20

C v (m/s)

t(s)029255

13

D v (m/s)

t(s)0221813

20

E v (m/s)

t(s)022185

20

14 The total distance travelled by the car is:A 440 m B 420 m C 350 mD 410 m E 400 m

15 The deceleration of the car, in m/s2, is equal to:A −5 B 4 C −4 D 5 E 2

Questions 16 and 17 refer to the following.The velocity–time graph, below, represents a particle moving in a straight line completing a distance of 150 metres.

v (m/s)

t(s)0 124 20

10

−5

16 At what time is the particle first at rest?A 4 s B 5 s C 6 s D 3 s E 25 s

17 The total time taken to travel the 150 metres is:A 25 s B 30 s C 22 sD 24 s E 28 s

18 An object which is initially at the origin moves in a straight line so that its velocity is v(x) = x + 1. The position, x, at any time, t, is given by:

A x = loge (t) B x = et + 1 C x t= −2 1D x = et − 1 E x = sin (t)

19 Given that dvdt v

=−1

2 4, v > 2; and that v = 3 when

t = 1 for an object moving in a straight line, the acceleration when t = 9 is:

A 14

B 12

C 16

D 112

E 23

20 A particle travels in a straight line with velocity v at time t.If the velocity of the particle is given by

vx

=−

2

1 2, for 0 < x < 1, then the acceleration is

given by:

A 2 sin−1(x) B 4

1

1

2

sin ( )−

−

x

xC 4

1 2 2

x

x( )−

D 2

1 2 2

x

x( )−E

2

1 232

x

x( )−

[©VCAA 2005]

exam Tip This was the worst-answered question in the Multiple choice section, with only 25% of students answering it correctly. A large percentage of students chose option E, which is the expression fordvdx

, whereas acceleration is given by vdvdx

. Students

also performed poorly on Question 30 of Part I in 2004, which similarly required knowledge of the various ways for expressing acceleration. This indicates that teachers and students need to give these ‘alternative forms’ more attention.

©VCAA Assessment report 2005

exTended reSponSe

1 Jogger A is running in a straight line at a constant speed of 4 m/s when passing jogger B who has stopped to tie a lace. Jogger B heads off in the same direction as A 6 seconds later, accelerating uniformly at 2 m/s2 until reaching a speed of 5 m/s.a Sketch a velocity–time graph showing the motion of both joggers.b How long is it after jogger A first passes jogger B until B catches up to A?c How far has jogger B travelled to catch jogger A?d How far ahead will jogger B be after jogger B has travelled 225 metres?

412

2 An object moving in a straight line has an acceleration of (x − 4). When x = 4, the velocity v = 0.a Explain why the velocity is decreasing from x = 0 to x = 4. b Hence, find v at x = 0.

3 The velocity of a particle moving in a straight line is given by:

v x x= − >4 02 ,

Find the acceleration when v = 3.

4 The velocity of a body moving in a straight line which is initially at the origin is given by:

v

t=

108

2cosπ

When t = 2, find:a the acceleration b the displacement from O.

5 A car travelling at 24 m/s overtakes a truck travelling at a constant speed of 17 m/s along a straight road.T seconds later, the car decelerates uniformly to rest.The truck continues at constant speed and it passes the car at the instant the car comes to a stop. This is exactly 60 seconds after the car had passed the truck.The velocity–time graph representing this situation is shown below.

T 60

17

0

24Car

Truck

v (m/s)

Find T.

6 Car A is 600 metres from the centre of the intersection when it starts from rest and accelerates uniformly at 4 m/s2, reaching a speed of 24 m/s which it maintains. At the instant car A takes off, car B is 780 metres from the centre of the intersection and travelling at a constant speed of 28 m/s. When car B is 52 metres from the centre of the intersection it decelerates uniformly at 5 m/s2.a Which car gets to the centre of the intersection first?b How far past the centre of the intersection is the first car when the second car reaches it? [©VCAA 2002]c If all other conditions remain the same, what constant acceleration would: i the second car need to have for a collision to occur? ii the first car need to have for a collision to occur?d If all other conditions remain the same, at what constant speed would: i the first car need to travel for a collision to occur? (Use a CAS

calculator or a numerical method to assist.) ii the second car need to travel for a collision to occur?

(Give answers correct to 2 decimal places where appropriate.)

7 A high diver jumps off a tower which is 44.1 metres above a deep tank of water. When the diver hits the water he is subject to an acceleration given by a = −0.4(v + 0.6)2 m/s2, v ≥ 0 where v is the velocity in the water at any time t seconds after impact with the water.a Find the velocity of the diver on impact with the water, assuming a

constant acceleration of 9.8 m/s2.b Find the velocity of the diver v(t) in the water at time t.c How long after the diver leaves the tower does he come to rest in the tank?d How far below the surface of the water does he descend?

exam Tip Some students had difficulty calculating the area-under-the-car graph. Others made simple algebraic or arithmetic slips, or had trouble with the time scale and used T + 60 as the total time. Many students unsuccessfully attempted to use constant acceleration formulas (e.g. putting u = 24, v = 0 and t = 60 for the car).


Car A

Car B

600 m

Centre ofintersection

780

m



8 A ball is dropped from the top of a building 35 m above the ground. At the same time, a second ball is thrown upwards with a speed of 20 m/s. The first ball is vertically above the second ball. The position above the ground of the first ball is given by x(t) = 35 − 4.9t2, while the position of the second ball above the ground is given by x(t) = 20t − 4.9t2.a At what time will the first ball have a speed that is the same as the speed of the second ball? (Careful;

speed is not the same as velocity.)b At the time the balls have the same speed, how far apart are they?c Eventually, the two balls collide. At what height above the ground do they collide?d Calculate the velocities of the two balls as they collide.In a second trial, the two balls are once again released. This time the second ball is thrown upwards with a speed V, such that when the first ball hits the second ball, the second ball is stationary.e Find the speed V and the time at which the balls collide.

9 An object has a velocity given by v x x x( ) ,= − ≥2 1 12 .

a Show that the acceleration of the object is 1 for all values of x ≥ 12.

b At t = 0, x = 1. Find the position of the object at all times t ≥ 0; that is, find x(t).

10 A motorcyclist passes a police car located at x = 0 at time t = 0. The speed of the motorcyclist is given by the equation v(t) = 15 + 0.2t and the acceleration of the police car is uniform at a rate of 2.0 m/s2 until the car reaches a speed of 25 m/s for t ∈ [0, 12.5], and 0 thereafter.a On the same set of axes, draw the velocity–time graphs for both the police car and the motorcyclist.b Determine the time(s) at which the two vehicles have the same speed.c Calculate the time(s) it takes for the police car to catch up to the motorcyclist and hence the distance(s)

travelled by the police car.

11 Martin is trying out his new spaceship by challenging a local alien to a race. He places his hand on the throttle and his spaceship starts to accelerate at a constant rate a for a time ∆t. During that time, ∆t, he travels a distance s through space. Martin’s spaceship has an initial speed u at the beginning of the time interval and a final speed v after a time ∆t. The local space police are on to Martin and want to photograph him just as he breaks the speed limit. However, they need to set their camera up at the precise point x where Martin is travelling at his average speed. The police know when the average speed will occur; it will occur at a time

Tt= ∆

2. They want to find out where the average speed will occur. Let x be the distance travelled when the

spaceship reaches its average velocity for the time interval ∆t. All distances and times are in metres (m) and seconds (s).It is clear that x is a fraction of s, the total distance covered during the time interval ∆t.a The initial velocity (u) is 0, the acceleration (a) is 1.0 and

the time interval (∆t) is 10. Find the value of x

and hence state the value of the fraction xs

.

b Find an equation for x in terms of only the acceleration, a; the initial speed, u; and s, the total distance travelled in the time interval ∆t.

c Under what conditions will xs=2

?

12 A body is released from rest and falls vertically (let a = 10 m/s2) for a distance of 100 m.a How far from the point of release will it be when it is moving at the average speed over that interval?b What is the average speed of the body for that 100-m interval?

13 When a projectile is fired from a body out into space, it decelerates due to the gravitational pull of the body.

The general equation for this deceleration is ak

r=

2 where r is the distance from the centre of the body and

k is a positive constant.

414

a On the surface of the moon (r = 1760 km), the deceleration is a = −1.6 m/s2. Find the value of k.b If a body is launched from the surface of the moon with a velocity of 500 m/s, fi nd the velocity of the

body at a distance r from the centre of the moon; that is, fi nd v(r).c Find the distance from the centre of the moon when the body is momentarily stationary.

14 A car travelling at 20 m/s passes a stationary police car, and then decelerates so that its velocity, v m/s, at time t seconds after passing the police car, is given by v = 20 − 2 tan−1 (t).

a After how many seconds will the car’s speed be 17 m/s? Give your answer correct to 1 decimal place.

b Explain why v will never equal 16.

c Write down a defi nite integral which gives the distance, x metres, travelled by the car after T seconds.

Three seconds later the police car starts to chase the passing car which has a polluting exhaust pipe. The police car accelerates so that its velocity, v m/s at time t seconds after the polluting car passed it, is given by

vt= −

−1313 2

71cos for t ∈ [3, 8].

d Write down an expression which gives how far the polluting car is ahead of the police car when t = 8 seconds. Find this distance in metres correct to 1 decimal place.

After accelerating for fi ve seconds the police car continues at a constant velocity.e At time t = Tc the police car catches the polluting car.

Write an equation which, when solved, gives the value of Tc.

f Find Tc correct to the nearest second.

[©VCAA 2007]

exam Tip The most common error was to attempt to solve 16 = 20 − 2 tan−1(t).


exam Tip Check the calculator mode, and round off to the correct number of decimal places.


exam Tip The major error involved either incorrect terminals or no terminals. Nearly all students who attempted this part did have at least dt present.


exam Tip This question was not well answered. Of those who attempted the question, the most common error was to have terminals in both integrals from 0 to 8.


exam Tip This question was not well done. Of the students who attempted this part, most calculated the constant speed of the police car, but many simply multiplied by Tc instead of Tc − 8 to fi nd the distance it travelled at constant speed.


exam Tip This question was not well done. Slightly more than half of those who set up a correct equation in Question 5e managed to arrive at the correct answer. Several methods could be used, each involving a numerical solution process.


[©VCAA 2007] [©VCAA 2007]

eBookpluseBookplus

Digital docTest YourselfChapter 8



eBookpluseBookplus aCTiViTieS

Chapter openerDigital doc

10 Quick Questions: Warm up with ten quick •questions on kinematics. (page 368)

8B Using antidifferentiationTutorial

We6 • int-0401: Watch a tutorial on how to calculate the velocity and displacement of a particle as a function of time given the acceleration as a function of time. (page 379)

8C Motion under constant accelerationTutorial

We8 • int-0402: Watch a tutorial on how to use constant acceleration formulas to calculate acceleration and distance travelled by a particle. (page 385)

8D Velocity–time graphsInteractivity

Velocity–time graphs • int-0349: Consolidate your understanding of the relationship between velocity and time. (page 391)

Tutorial

We 12 • int-0403: Watch a tutorial on how to sketch a velocity–time graph and use it to calculate distance travelled. (page 393)

Digital doc

WorkSHEET 8.1: Make calculations about bodies •and particles in motions with respect to time, displacement, velocity and acceleration. (page 399)

8E Applying differential equations torectilinear motion

Interactivity

We 16 • int-0404: Calculate the velocity, acceleration and position as functions of time, given a differential equation. (page 402)

Digital doc

WorkSHEET 8.2: Make calculations about bodies •and particles in motion with respect to time, displacement, velocity and acceleration, and sketch velocity–time graphs. (page 407)The falling ball bearing: Investigate the kinematics •of a falling ball bearing. (page 407)

Chapter reviewDigital doc

Test Yourself: Take the end-of-chapter test to test •your progress. (page 414)

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