ch2-kinematics

7/28/2019 Ch2-Kinematics

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CHAPTER 2 Kinematics

Question 2.1

The driver of a car suddenly brakes when he sees the van in front of him stop. If the carinitially is moving at 30 m s1 and the maximum deceleration of this car is7.5 m s2, calculate the distance travelled by the car before it comes to a stand-still(a) if reaction time is ignored, (b) if the drivers reaction time is 0.5 s.

Solution 2.1

(a) v2

= u2 + 2as

0 = (30)2 + 2(7.5) s s = 60 m

(b) Distance travelled during reaction time = (30)(0.5) m = 15 mTotal distance travelled = (15 + 60) m = 75 m

Question 2.2

The variation of acceleration with time for a body is as shown in the graph below.Assuming that the body starts from rest,

(a) plot a graph of velocity against time for the motion of the body.(b) determine the total distance travelled during the first 8 s.

Solution 2.2

(a)

(b) Total distance travelled equals to the area under the graph.

Total distance travelled = 135 m

t/s

40

60

20

20

50

40

60

1 2 3 4 5 6 7 8

15

3

45

v/m s1

a/m s2

t/s

20

10

10

15

20

1 2 3 4 5 6 7 8 9

v1

= u + at= 0 + (15)3= 45 m s1

v2

= 45 + 20(2)= 5 m s1

v3

= 5 + (10)(2)= 15 m s1


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Question 2.3

A bullet is fired vertically upwards with an initial velocity of 300 m s1. Assuming that airresistance can be reflected and the acceleration due to gravity is 9.81 m s 2,

(a) find the maximum height reached by the bullet; and(b) the time elasped before the bullet falls back to the level of firing.

Solution 2.3

(a) Using v2 = u2 + 2as0 = (300)2 + 2(9.81)s

s = 4.59 103 m

(b) Let t1

be the time to reach maximum heightv = u + at

1

0 = 300 + (9.81)t1

t1

= 30.6 s

Since the condition experienced by the bullet going upwards and falling downwards

are the same, Total time taken = 2(30.6) s

= 61.2 s

Question 2.4

A rocket is launched vertically upwards. The thrust of the engine provides a uniform

acceleration of 400 m s2 for 10 s until all the fuel is burnt. It then falls freely back to theground. Calculate(a) the maximum speed of the rocket,

(b) the time the rocket took to reach maximum height; and

(c) the maximum height reached.Assume g

0= 10 m s2

Solution 2.4

(a) Using v = u + atv = 0 + (400)(10)

= 4.0 103 m s1

(b) At the end of the first 10 s, the rocket continues to move upwards under the action ofgravity.Using v = u + at

0 = 4.0 103

+ (10)t t = 400 s

Total time taken to reach maximum height = 400 + 10= 410 s

1(c) Distance travelled in the first 10 s is calculated from s = ut+ at22

1 s

1= 0 + (400)(10)2

2

= 2.0 104 m

Distance travelled after the fuel has burnt out = s2

v2 = a2 + 2as

0 = (4.0 103)2 2(10.0)s s = 8.0 105 m

Maximum height reached = s1

+ s2

= 82 104 m


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Question 2.6

The maximum speed that a football can be kicked by a player is 25 m s 1. Neglecting airresistance, find the maximum horizontal distance travelled by the ball before it lands onthe ground.

Solution 2.6

(a) For maximum horizontal range = 45 and 2 = 90

u2 sin 2Using R =

g

252 (sin 90)=

(9.81)

= 63.7 m

Question 2.5

An arrow which has been fired vertically upwards leaves the bow at a speed of 30.0 m s1.(a) What is the height of the arrow above the bow when it has been in flight for 2.0 s?

(b) How long does it take the arrow to return to the same level as the bow?Take g = 9.81 m s2

Solution 2.5

1(a) From s = at+ at2

21

= (30.0)(2.0) + (9.81)(2.0)22

= 40.4 m

At time = 2.0 s, the arrow is 40.4 m above the bow.

(b) Use v = u + atWhen the arrow returns to the same level as the bow, its velocity is 30.0 m s1

downwards.

30.0 = 300 + (9.81) tt = 6.1 s

Question 2.7

A football is kicked with an initial speed of 30 m s1 at an angle of 60 to the ground

towards a player standing 100 m away. Calculate(a) the total time the football is in the air,(b) the horizontal distance travelled by the ball.

If the player starts to run towards the ball at the instant it was kicked, what is theminimum constant speed he must run in order to get the ball before it hit the ground?

Solution 2.7

2u sin (a) Total time of flight T =

g(2)(30)(sin 60)

= = 5.3 s(9.81)


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Question 2.8

A stone is thrown with an initial horizontal speed of 20.0 m s 1 over a cliff of height75 m. Calculate the total time the stone is in the air, and the horizontal distance betweenthe foot of the cliff and the point where the stone lands.Another stone is thrown with the same initial speed but at an angle to the ground. Findthe furthest point from the cliff that the stone can be thrown so that it just clears the edge

of the cliff. How far is the point where the stone lands from the foot of the cliff?

Solution 2.8

Consider the vertical component of motion

1s = ut+ at2

2

175 = 0 + (9.81)t2

2

t = 3.91 s

Total time the stone is in the air is 3.91 sx = ut = (20.0)(3.91) m

= 78.2 m

For maximum horizontal range = 45

u 2 sin 2 R =

g

(202)(sin 90)= = 40.8 m

9.81

(b) Horizontal distance travelled = u cos 60T= (30)(cos 60)(5.3)

= 79.5 m

Distance ran by player = (100 79.5) = 20.5 m

20.5v = m s1

5.3

= 3.9 m s1

75 m

x

u= 20.0 m s1

R

x1

u= 20.0 m s1

45

45

75 m

When clearing the edge of the cliff, the velocity of the stone is 20.0 m s1, with an angle of45 to the horizontal.At this instant, horizontal component of velocity = 20 cos 45

Vertical component of velocity = 20 sin 45Let the time taken from the edge of the cliff to the foot of the cliff be t

1

First we need to find the final vertical component of velocity of the stone just beforehitting the ground.

Using v2 = u2 + 2asv2 = (20 sin 45)2 + 2(9.81)(75)v = 35.7 m s1

Since direction is downwards, v= 35.7 m s1

From v = a + at35.7 = 20 sin 45 + (9.8)t1

t1

= 2.2 s

x1

= (u cos 45)t1

= (20 cos 45)(2.2)= 31.1 m

60

uv

79.5 m

100 m

20.5

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