Solutionbank M1 - Webflow

Solutionbank M1Heinemann Modular Maths for Edexcel AS and A-level

2 Kinematics in one dimensionExercise B, Question 1

Question:

A car accelerates at 2 m s − 2 from rest for 10 seconds along a straight road.

(a) Find the distance travelled by the car and the speed it reaches.

(b) After the 10 seconds its acceleration changes to 0.5 m s − 2 and then remains constant for a further 5 seconds. Find the speed of the car and the total distance that it has travelled at the end of the 15 seconds.

Solution:

(a)

(b)

∴ total distance = s1 + s2 = 100 + 106.25 = 206.25 metres

© Harcourt Education Ltd 2005

s = ut + at21

2

s1 = 0 × 10 + × 2 × 1021

2

s1 = 100 metres

v = u + atv1 = 0 + 2 × 10

v1 = 20 m s − 1

v = u + atv2 = 20 + 0.5 × 5

v2 = 22.5 m s − 1

s = ut + at21

2

s2 = 20 × 5 + × 0.5 × 52 = 106.25 m1

2

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Question:

A car accelerates uniformly from 10 m s − 1 to 20 m s − 1 as it travels 500 m along a straight road.

(a) Find the acceleration of the car.

(b) Find the time taken by the car to travel the 500 m.

Solution:

(a)

(b)


v2 = u2 + 2as

202 = 102 + 2 × a × 500400 = 100 + 1000a400 − 100 = 1000a

∴ 300 = 1000a

∴ a = = 0.3 m s − 2300

1000

s = t( u + v )

2

500 = × t1( 10 + 20 )

2

∴ 500 = 15 × t1

t1 = = 33 seconds500

15

1

3

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Question:

A car accelerates uniformly as it moves along a straight road. Its velocity increases from 5 m s − 1 to 12 ms − 1 in a 10-second period.

(a) Find the acceleration of the car.

(b) Find the distance travelled by the car.

Solution:

(a)

(b)

or


v = u + at12 = 5 + a × 1012 − 5 = 10a

a = = 0.7 m s − 27

10

v2 = u2 + 2as

122 = 52 + 2 × 0.7 × s

∴ 144 − 25 = 1.4s

s = = 85 m119

1.4

s = t( u + v )

2

s1 = × 10( 5 + 12 )

2

s1 = 85 m

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Question:

A lift rises vertically from rest, accelerating at a constant rate until it reaches a speed of 1.6 m s − 1 after 8 seconds.

(a) Find the acceleration of the lift.

(b) The lift continues to accelerate for a further 2 seconds. Find the distance that the lift has now risen.

(c) The lift then slows down, at a constant rate, and stops after a further 5 seconds. Find the total distance travelled by the lift.

Solution:

(a)

(b)

v = u + at1.6 = 0 + a × 8

∴ a = 0.2 m s − 2

s = . t in 1st part( u + v )

2

s1 = × 8( 0 + 1.6 )

2

s1 = 6.4 m

s = ut + at2 in 2nd part1

2

s2 = 1.6 × 2 + × 0.2 × 221

2

s2 = 3.6 m

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∴ total distance = s1 + s2 = 6.4 + 3.6 = 10 metres

(c)


In second part, v = u + atv2 = 1.6 + 0.2 × 2 = 2 m s − 1

In last part, s = × t( u + v )

2

s3 = × 5 = 5 metres( 2 + 0 )

2

∴ total distance = 10 + 5

= 15 metres

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Question:

A car accelerates uniformly from a speed of 50 km h − 1 to a speed of 80 km h − 1 in 20 seconds as it travels along a straight road.

(a) Calculate the acceleration of the car in m s − 2.

(b) Calculate the distance travelled during the 20 seconds, giving your answer to the nearest metre.

Solution:

(a)

(b) s = × t


50 km h − 1 = m s − 150 × 1000

3600 = 13 m s − 18

9

80 km h − 1 = m s − 180 × 1000

3600= 22 m s − 12

9

v = u + at

222

9= 13 + a × 20

8

9

∴ 22 − 132

9

8

9= 20a

∴ 81

3

20

= a

∴ a = m s − 2 = 0.416 m s − 2 or 0.417 m s − 2 (3 s.f.)5

12

( u + v )

2

s1 = × 20( 13 + 22 )

8

9

2

9

2

s1 = 361 m = 361 metres (nearest metre)1

9

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Question:

A van travelling at 40 mph skids to a halt in a distance of 15 m. Find the acceleration of the van and the time taken to stop, assuming that the deceleration is uniform and the van travels along a straight line. (Assume 1 mile = 1600 m.)

Solution:

40mph = m s − 1 = 17 m s − 1


40 × 1600

3600

7

9

v2 = u2 + 2as

02 = ( 17 ) 2 + 2 ( − r ) ( 15 )7

9

∴ 30 r = 3164

81

r = 10 = 10.5 m s − 2 (3 s.f.), deceleration.130

243

s = × t( u + v )

2

15 = × t( 17 + 0 )

7

9

2

∴ 2 × 15 = 17 × t7

9

∴2 × 15

177

9

= t

∴ t = 1 s11

16

= 1.69 s (3 s.f.)

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Question:

A train signal is placed so that a train can decelerate uniformly from a speed of 96 km h − 1 to come to rest at the end of a platform. For passenger comfort the deceleration must be no greater than 0.4 m s − 2. Assume that the train travels along a straight line. Calculate:

(a) the shortest distance the signal can be from the end of the platform

(b) the shortest time for the train to decelerate.

Solution:

96 km h − 1 = m s − 1 = 26 m s − 1

(a) v2 = u2 + 2as with maximum deceleration

(b)


96 × 1000

3600

2

3

02 = ( 26 ) 2 + 2 ( − 0.4 ) ( s1 )2

3

∴ 0.8s1 = ( 26 ) 22

3

∴ s1 = = 888.88... m( 26 ) 22

3

0.8

i.e. s1 = 889 m (3 s.f.)

v = u + at

0 = 26 + ( − 0.4 ) × t12

3

∴ 0.4t1 = 262

3

t1 = = 66.666... = 66.7 s (3 s.f.)26

2

3

0.4

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Question:

A rocket is travelling with a velocity of 80 m s − 1. The engines are switched on for 6 seconds and the rocket accelerates uniformly at 40 m s − 2. Assume that the rocket is always moving in the same direction. Calculate the distance travelled over the 6 seconds.

Solution:


s = ut + at21

2

s1 = 80 × 6 + × 40 × 621

2

s1 = 1200 m

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Question:

The world record for the men's 60 m race was 6.41 seconds.

(a) Assuming that the race was carried out under constant acceleration, calculate the acceleration of the runner and his speed at the end of the race.

(b) Now assume that in a 100 m race the runner accelerates for the first 60 m and completes the race by running the next 40 m at the speed you calculated in (a).Calculate the time for the athlete to complete the race.

Solution:

(a)

s = × t

60 = × 6.41 , ∴ = v

∴ v = 18.720... = 18.7 m s − 1 (3 s.f.)

(b) uniform speed so

∴ total time = 6.41 + 2.1366 = 8.55 s (3 s.f.)

s = ut + at21

2

60 = 0 × 6.41 + × a × 6.4121

2

∴60

× 6.4121

2

= a

i.e. a = 2.9205... = 2.92 m s − 2 (3 s.f.)

( u + v )

2

( 0 + v )

2

60 × 2

6.41

s = ut

40 = 18.720 × t2∴ t2 = 2.1366... s

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Question:

The world record for the men's 100 m sprint was 9.83 seconds. Assume that the last 40 m was run at constant speed and that the acceleration during the first 60 m was constant.

(a) Calculate this speed.

(b) Calculate the acceleration of the athlete.

Solution:

(a) × t1 × v = 60 → t1 =


1

2

120

v

( 9.83 − t1 ) × v = 40

∴ ( 9.83 − ) × v120

v= 40

∴ 9.83v − 120 = 40

∴ 9.83v = 40 + 120 , i.e. 9.83v = 160

∴ v = = 16.276... = 16.3 m s − 1 (3 s.f.)160

9.83

(b) acceleration =v

t1

but × t1 × 16.276...1

2= 60 from the first equation

∴ t1 =60 × 2

16.276...

t1 = 7.3725 s

∴ acceleration =16.276...

7.3725

= 2.21 m s − 2 (3 s.f.)

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Question:

Telegraph poles, 40 m apart stand alongside a straight railway line. The time taken for a locomotive to pass the two gaps between three consecutive poles are 2.5 seconds and 2.3 seconds, respectively. Assume that the acceleration of the train is constant. Calculate the acceleration of the train and the speed past the first post.

Solution:

s = ut + at2


1

2

{

to 1st post 40 = u × 2.5 + × a × 2.521

2

to 2nd post 80 = u × 4.8 + × a × 4.821

2

∴ 4.8 × 40 = ( u × 2.5 + 3.125a ) × 4.8

2.5 × 80 = ( u × 4.8 + 11.52a ) × 2.5

{ subtract∴ 192 = 12u + 15a

200 = 12u + 28.8a

8 = 13.8a

∴ a = = 0.57971... = 0.580 m s − 2 (3 s.f.)8

13.8

then 40 = 2.5u + 3.125 × 0.57971...

∴ 40 − 3.125 × 0.57971... = 2.5u

u = = 15.3 m s − 1 (3 s.f.)[ 40 − 3.125 × 0.57971... ]

2.5

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Question:

A sprinter starts from rest, and accelerates at 2 m s − 2 for the first 4 seconds of a race. Assume that the sprinter moves along a straight line.

(a) Find the distance travelled by the sprinter in the first 4 seconds.

(b) Find the speed of the sprinter at the end of the first 4 seconds.

(c) The sprinter then travels at this speed for the remainder of the race. He travels a total distance of 100 metres. Find the total time that he takes to complete the race. [A]

Solution:

(a)

(b)

(c)

∴ total time = 4 + 10.5 = 14.5 seconds


s = ut + at21

2

s1 = 0 × 4 + × 2 × 421

2

s1 = 16 m

v = u + atv1 = 0 + 2 × 4

v1 = 8 m s − 1

In final part, s = ut

84 = 8 × t3 , ∴ t3 = = 10.5 s84

8

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Question:

A cyclist is travelling along a straight horizontal road. As she passes a bus stop she sees a red traffic light ahead of her. She continues to travel with a constant speed of 3 m s − 1 for 20 seconds and then decelerates at a constant rate of 0.2 m s − 2 until coming to rest at the traffic light.

(a) Calculate the distance between the bus stop and the traffic light.

(b) Calculate the time the cyclist takes to travel from the bus stop to the traffic light.

(c) Calculate the average speed of the cyclist between the bus stop and the traffic light. [A]

Solution:

(a)

∴ total distance = s1 + s2 = 60 + 22.5 = 82.5 metres

(b) 2nd part, v = u + at 0 = 3 + ( − 0.2 ) t2 , ∴ 0.2t2 = 3

∴ t2 = = 15 seconds

∴ total time = 20 + 15 = 35 seconds

(c)


1st part, s = uts1 = 3 × 20 = 60 m

2nd part, v2 = u2 + 2as

02 = 32 + 2 ( − 0.2 ) ( s2 )

∴ 0.4s2 = 32

s2 = = 22.5 m32

0.4

3

0.5

Average speed = = = 2.3571...total distance

total time

82.5

35

= 2.36 m s − 1 (3 s.f.)

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Question:

A set of traffic lights covers road repairs on a road. The traffic lights are 80 m apart. Assume that a car accelerates at 2 m s − 2 from rest until it reaches a speed of 10 m s − 1 and then travels at a constant speed. What is the least time taken for a car starting from rest at the first set of lights to reach the next set?

Solution:

1st part, v2 = u2 + 2as

∴ total time = t1 + t2 = 5 + 5.5 = 10.5 seconds


102 = 02 + 2 ( 2 ) ( s1 )

100 = 4s1

∴ s1 = 25 m

1st part, v = u + at

10 = 0 + 2 × t1∴ t1 = 5 s

2nd part s2 = 80 − 25 = 55 m

∴ s = ut 55 = 10 × t2

∴ t2 = = 5.5 s55

10

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Question:

A train starts from rest and moves along a straight track with constant acceleration m s − 2 for 2 minutes.

For the next 4 minutes the train moves with zero acceleration, after which a uniform retardation of 2 m s − 2

brings it to rest. Find the total distance travelled by the train from starting to stopping. [A]

Solution:


1

3

1st part, s = ut + at21

2

s1 = 0 × 120 + × × 12021

2

1

3

s1 = 2400 m

v = u + at

v1 = 0 + × 1201

3

v1 = 40 m s − 1

2nd part, s = uts2 = 40 × 240 = 9600 m

3rd part, v2 = u2 + 2as

02 = 402 + 2 ( − 2 ) s3

∴ 4s3 = 1600

s3 = 400 m

∴ total distance = s1 + s2 + s3 = 2400 + 9600 + 400

= 12400 metres

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Question:

As a train leaves a station it accelerates, from rest, at 0.8 m s − 2 for 30 seconds, travels at a constant speed for the next 5 minutes and then slows down, stopping in 20 seconds at a second station. Assume that the train moves along a straight track.

(a) Find the maximum speed of the train.

(b) Find the distance travelled by the train between the stations, clearly stating any assumptions that you have made. [A]

Solution:

(a) 1st part, v = u + at

(b) 1st part, s = ut + at2

s1 = 0 × 30 + × 0.8 × 302 = 360 metres

2nd part, s = uts2 = 24 × 300 = 7200 metres


v1 = 0 + 0.8 × 30

v1 = 24 m s − 1

1

2

1

2

3rd part, s = t( u + v )

2

s3 = × 20 = 240 metres( 24 + 0 )

2

∴ total distance = s1 + s2 + s3 = 360 + 7200 + 240

= 7800 metresThe assumptions are • the acceleration in the first part of the motion remains constant

• the deceleration in the final part of the motion is constant.

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Question:

As a lift moves upwards from rest it accelerates at 0.8 m s − 2 for 2 seconds, then travels 4 m at constant speed and finally slows down, with a constant deceleration, stopping in 3 seconds.Find the total distance travelled by the lift and the total time taken. [A]

Solution:

and total time = t1 + t2 + t3 = 2 + 2.5 + 3 = 7.5 seconds

1st part, s = ut + at21

2

s1 = 0 × 2 + × 0.8 × 22 = 1.6 m1

2

v = u + atv1 = 0 + 0.8 × 2 = 1.6 m s − 1

2nd part, s = ut

4 = 1.6 × t2

∴ t2 = = 2.5 seconds4

1.6

3rd part, s = t( u + v )

2

s3 = × 3 = 2.4 m( 1.6 + 0 )

2

∴ total distance = s1 + s2 + s3

= 1.6 + 4 + 2.4= 8 metres.

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Question:

Two cars, A and B, are initially at rest side by side. A starts off on a straight track with an acceleration of 2 ms − 2. Five seconds later B starts off on a parallel track to A, with acceleration 3.125 m s − 2.

(a) Calculate the distance travelled by A after 5 seconds.

(b) Calculate the time taken for B to catch up with A.

(c) Find the speeds of A and B at that time.

Solution:

(a)

(b)

when B catches A, s3 = s1 + s2

.

For A , s = ut + at21

2

s1 = 0 × 5 + × 2 × 521

2

s1 = 25 metres

1st part, A , v = u + atv1 = 0 + 2 × 5 = 10 m s − 1

2nd part, A , s = ut + at21

2

s2 = 10t2 + × 2 × t221

2

i.e. s2 = 10t2 + t22

B , s = ut + at21

2

s3 = 0 × t2 + × 3.125 × t221

2

i.e. s3 = 1.5625t22

i.e. 1.5625t22 = 25 + ( 10t2 + t2

2 )

i.e. 0.5625t22 − 10t2 − 25 = 0

∴ t2 =− ( − 10 ) ± \ ( − 10 ) 2 − 4 ( 0.5625 ) ( − 25 )

2 × 0.5625

∴ t2 = 20 or t2 = − 2.222... but time can't be negative

∴ t2 = 20 seconds

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(c)


A, 2nd part, v = u + at vA = 10 + 2 × 20 = 50 m s − 1

B , v = u + at vB = 0 + 3.125 × 20 = 62.5 m s − 1

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Question:

A cyclist sets off from rest and moves along a straight horizontal road until she again comes to rest. The motion of the cyclist can be modelled as three separate stages.In the first stage she accelerates uniformly from rest for 5 seconds until she reaches a velocity V m s − 1.She then moves with constant velocity V m s − 1 for 55 seconds. Finally she moves with a constant retardation for 10 seconds until coming to rest.

(a) Sketch a velocity-time graph to show the motion of the cyclist.

(b) Given that the total length of the journey is 300 m, find the value of V. [A]

Solution:

(b) Using distance is the area under the graph:


[ × 5 × V ] + [ 55 × V ] + [ × 10 × V ]1

2

1

2= 300

∴ 2.5V + 55V + 5V = 300

62.5V = 300

V =300

62.5

V = 4.8 m s − 1

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Question:

Two cars are initially 36 m apart travelling in the same direction along a straight, horizontal road. The car in front is initially travelling at 10 m s − 1, but decelerating at 2 m s − 2. The other car travels at a constant 15 m s − 1.

(a) Model the cars as particles. By finding the distance travelled by each car after t seconds, show that the distance between the two cars is 36 − 5t − t2 metres. Find when they would collide if neither car takes avoiding action.

(b) Would it be necessary to revise your answers to part (a) if the cars were not modelled as particles? Give reasons to support your answer. [A]

Solution:

(a)

(b) No, as vehicles are 36 m apart.

1st car, s = uts1 = 15t

2nd car, s = ut + at21

2

s2 = 10t + ( − 2 ) t21

2

s2 = 10t − t2

∴ distance between them = 36 + s2 − s1

= 36 + ( 10t − t2 ) − ( 15t )

= 36 − 5t − t2 metres.

∴ when collide, 36 − 5t − t2 = 0

i.e. 0 = t2 + 5t − 360 = ( t + 9 ) ( t − 4 )i.e. t = − 9 but can't have negative time, or t = 4 seconds

∴ t = 4 seconds

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