SOLUTIONS TO CONCEPTS CHAPTER – 3


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SOLUTIONS TO CONCEPTS
CHAPTER – 3

1. a) Distance travelled = 50 + 40 + 20 = 110 m b) AF = AB – BF = AB – DC = 50 – 20 = 30 M His displacement is AD

N W

B E 50 m

40 m 40 m

C 20 m D

AD = AF2  DF2  302  402  50m
In AED tan  = DE/AE = 30/40 = 3/4   = tan–1 (3/4)

30 m

S

A 

E

A  Initial point

(starting point)

His displacement from his house to the field is 50 m, tan–1 (3/4) north to east. 2. O  Starting point origin. i) Distance travelled = 20 + 20 + 20 = 60 m ii) Displacement is only OB = 20 m in the negative direction.

Y

B

A

(–20 m, 0) O

X (20 m, 0)

Displacement  Distance between final and initial position.

3. a) Vave of plane (Distance/Time) = 260/0.5 = 520 km/hr.

b) Vave of bus = 320/8 = 40 km/hr.

c) plane goes in straight path

velocity = Vave = 260/0.5 = 520 km/hr.

d) Straight path distance between plane to Ranchi is equal to the displacement of bus.

 Velocity = Vave = 260/8 = 32.5 km/hr.

4. a) Total distance covered 12416 – 12352 = 64 km in 2 hours.

Speed = 64/2 = 32 km/h

b) As he returns to his house, the displacement is zero.

Velocity = (displacement/time) = 0 (zero).

5. Initial velocity u = 0 ( starts from rest)

Final velocity v = 18 km/hr = 5 sec

(i.e. max velocity)

Time interval t = 2 sec.

 Acceleration = aave = v  u  5 = 2.5 m/s2.
t2

6. In the interval 8 sec the velocity changes from 0 to 20 m/s.

Average acceleration = 20/8 = 2.5 m/s2  change in velocity 



time



Distance travelled S = ut + 1/2 at2

 0 + 1/2(2.5)82 = 80 m.

7. In 1st 10 sec S1 = ut + 1/2 at2  0 + (1/2 × 5 × 102) = 250 ft.

20

Initial velocity

10

u = 0

48 Time in sec

At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec.  From 10 to 20 sec (t = 20 – 10 = 10 sec) it moves with uniform velocity 50 ft/sec,
3.1

1000
750 S (in ft)
250

0

10 20 30

t (sec)

Chapter-3

Distance S2 = 50 × 10 = 500 ft Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s2. At 20 sec velocity is 50 ft/sec.

t = 30 – 20 = 10 s S3 = ut + 1/2 at2
= 50 × 10 + (1/2)(–5)(10)2 = 250 m

Total distance travelled is 30 sec = S1 + S2 + S3 = 250 + 500 + 250 = 1000 ft.

8. a) Initial velocity u = 2 m/s.

final velocity v = 8 m/s

time = 10 sec,

acceleration = v  u  8  2 = 0.6 m/s2
ta 10
b) v2 – u2 = 2aS

 Distance S = v2  u2 = 82  22 = 50 m.

2a

2  0.6

t 8 6 4 2
5

10 t

c) Displacement is same as distance travelled.

Displacement = 50 m.

9. a) Displacement in 0 to 10 sec is 1000 m.

time = 10 sec.

100

Vave = s/t = 100/10 = 10 m/s. b) At 2 sec it is moving with uniform velocity 50/2.5 = 20 m/s.
at 2 sec. Vinst = 20 m/s. At 5 sec it is at rest.

50

0 2.5 5 7.5 10

t 15

(slope of the graph at t = 2 sec)

Vinst = zero.

At 8 sec it is moving with uniform velocity 20 m/s

Vinst = 20 m/s

At 12 sec velocity is negative as it move towards initial position. Vinst = – 20 m/s.

10. Distance in first 40 sec is,  OAB + BCD

= 1 × 5 × 20 + 1 × 5 × 20 = 100 m.

2

2

Average velocity is 0 as the displacement is zero.

5 m/s A

O

B

20

D t (sec) 40

11. Consider the point B, at t = 12 sec At t = 0 ; s = 20 m

and t = 12 sec s = 20 m

20

So for time interval 0 to 12 sec

10

Change in displacement is zero.

So, average velocity = displacement/ time = 0

 The time is 12 sec.

12. At position B instantaneous velocity has direction along BC . For y

average velocity between A and B.

4

Vave = displacement / time = (AB / t) t = time

2

C B

10 12

20

B C

3.2

x

2

4

6

We can see that AB is along BC i.e. they are in same direction. The point is B (5m, 3m). 13. u = 4 m/s, a = 1.2 m/s2, t = 5 sec Distance = s = ut  1 at2
2
= 4(5) + 1/2 (1.2)52 = 35 m. 14. Initial velocity u = 43.2 km/hr = 12 m/s
u = 12 m/s, v = 0 a = –6 m/s2 (deceleration) Distance S = v2  u2 = 12 m
2(6)

Chapter-3

3.3

Chapter-3

15. Initial velocity u = 0 Acceleration a = 2 m/s2. Let final velocity be v (before applying breaks)

t = 30 sec

v = u + at  0 + 2 × 30 = 60 m/s

a) S1 = ut  1 at2 = 900 m
2

when breaks are applied u = 60 m/s

v = 0, t = 60 sec (1 min) Declaration a = (v – u)/t = = (0 – 60)/60 = –1 m/s2.

S2 = v2  u2 = 1800 m
2a

Total S = S1 + S2 = 1800 + 900 = 2700 m = 2.7 km. b) The maximum speed attained by train v = 60 m/s

c) Half the maximum speed = 60/2= 30 m/s

Distance S = v2  u2 = 302  02 = 225 m from starting point

2a

22

When it accelerates the distance travelled is 900 m. Then again declarates and attain 30 m/s.  u = 60 m/s, v = 30 m/s, a = –1 m/s2

Distance = v2  u2 = 302  602 = 1350 m

2a

2(1)

Position is 900 + 1350 = 2250 = 2.25 km from starting point.

16. u = 16 m/s (initial), v = 0, s = 0.4 m.

Deceleration a = v2  u2 = –320 m/s2.
2s

Time = t = v  u  0 16 = 0.05 sec.
a  320

17. u = 350 m/s, s = 5 cm = 0.05 m, v = 0

Deceleration = a = v2  u2 = 0  (350)2 = –12.2 × 105 m/s2.

2s

2  0.05

Deceleration is 12.2 × 105 m/s2.

18. u = 0, v = 18 km/hr = 5 m/s, t = 5 sec

a = v  u  5  0 = 1 m/s2.

t

5

s = ut  1 at2 = 12.5 m
2

a) Average velocity Vave = (12.5)/5 = 2.5 m/s. b) Distance travelled is 12.5 m.

19. In reaction time the body moves with the speed 54 km/hr = 15 m/sec (constant speed)

Distance travelled in this time is S1 = 15 × 0.2 = 3 m.
When brakes are applied, u = 15 m/s, v = 0, a = –6 m/s2 (deceleration)

3.4

S2 = v2  u2  0 152 = 18.75 m

2a

2(6)

Total distance s = s1 + s2 = 3 + 18.75 = 21.75 = 22 m.

Chapter-3

3.5

Chapter-3

20.

Driver X Reaction time 0.25

Driver Y Reaction time 0.35

A (deceleration on hard braking = 6 m/s2)

Speed = 54 km/h Braking distance a= 19 m Total stopping distance b = 22 m

Speed = 72 km/h Braking distance c = 33 m Total stopping distance d = 39 m.

B (deceleration on hard braking = 7.5 m/s2)

Speed = 54 km/h Braking distance e = 15 m Total stopping distance f = 18 m

Speed = 72 km/h Braking distance g = 27 m Total stopping distance h = 33 m.

a = 02 152 = 19 m
2(6)

So, b = 0.2 × 15 + 19 = 33 m

Similarly other can be calculated.

Braking distance : Distance travelled when brakes are applied.

Total stopping distance = Braking distance + distance travelled in reaction time.

21. VP = 90 km/h = 25 m/s.

VC = 72 km/h = 20 m/s. In 10 sec culprit reaches at point B from A.

Police t=0

t = 10 sec AP

Distance converted by culprit S = vt = 20 × 10 = 200 m.
At time t = 10 sec the police jeep is 200 m behind the culprit.

culprit

BC

Time = s/v = 200 / 5 = 40 s. (Relative velocity is considered).

In 40 s the police jeep will move from A to a distance S, where

S = vt = 25 × 40 = 1000 m = 1.0 km away.

 The jeep will catch up with the bike, 1 km far from the turning.

22. v1 = 60 km/hr = 16.6 m/s.

v2 = 42 km/h = 11.6 m/s.

Relative velocity between the cars = (16.6 – 11.6) = 5 m/s. Distance to be travelled by first car is 5 + t = 10 m. Time = t = s/v = 0/5 = 2 sec to cross the 2nd car. In 2 sec the 1st car moved = 16.6 × 2 = 33.2 m
H also covered its own length 5 m.

V1  5 m

V2  5 m

Before crossing

V2 

10 m

V1 

After crossing

 Total road distance used for the overtake = 33.2 + 5 = 38 m. 23. u = 50 m/s, g = –10 m/s2 when moving upward, v = 0 (at highest point).

a) S = v2  u2  0  502 = 125 m
2a 2(10)

maximum height reached = 125 m
b) t = (v – u)/a = (0 – 50)/–10 = 5 sec c) s = 125/2 = 62.5 m, u = 50 m/s, a = –10 m/s2,

3.6

v2 – u2 = 2as

Chapter-3

 v = (u2  2as)  502  2(10)(62.5) = 35 m/s.

24. Initially the ball is going upward u = –7 m/s, s = 60 m, a = g = 10 m/s2
s = ut  1 at2  60 = –7t + 1/2 10t2
2
 5t2 – 7t – 60 = 0

t = 7  49  4.5(60) = 7  35.34

25

10

taking positive sign t = 7  35.34 = 4.2 sec ( t  –ve)
10

Therefore, the ball will take 4.2 sec to reach the ground. 25. u = 28 m/s, v = 0, a = –g = –9.8 m/s2

a) S = v2  u2  02  282 = 40 m

2a

2(9.8)

b) time t = v  u  0  28 = 2.85
a  9.8

t = 2.85 – 1 = 1.85

v = u + at = 28 – (9.8) (1.85) = 9.87 m/s.

 The velocity is 9.87 m/s.

c) No it will not change. As after one second velocity becomes zero for any initial velocity and deceleration is g = 9.8 m/s2 remains same. Fro initial velocity more than 28 m/s max height increases.
26. For every ball, u = 0, a = g = 9.8 m/s2

 4th ball move for 2 sec, 5th ball 1 sec and 3rd ball 3 sec when 6th ball is being dropped.

For 3rd ball t = 3 sec

6th

S3 = ut  1 at2 = 0 + 1/2 (9.8)32 = 4.9 m below the top.

5th 4th

2

3rd

For 4th ball, t = 2 sec

S2 = 0 + 1/2 gt2 = 1/2 (9.8)22 = 19.6 m below the top (u = 0)

For 5th ball, t = 1 sec

S3 = ut + 1/2 at2 = 0 + 1/2 (9.8)t2 = 4.98 m below the top.

27. At point B (i.e. over 1.8 m from ground) the kid should be catched.

For kid initial velocity u = 0 Acceleration = 9.8 m/s2
Distance S = 11.8 – 1.8 = 10 m

10m 11.8

S = ut  1 at2  10 = 0 + 1/2 (9.8)t2

1.8m

2

 t2 = 2.04  t = 1.42.

7m

In this time the man has to reach at the bottom of the building.

Velocity s/t = 7/1.42 = 4.9 m/s.

28. Let the true of fall be ‘t’ initial velocity u = 0

3.7

Acceleration a = 9.8 m/s2

Chapter-3

Distance S = 12/1 m

 S = ut  1 at2
2
 12.1 = 0 + 1/2 (9.8) × t2

1.66 m/s

 t2 = 12.1 = 2.46  t = 1.57 sec

2.6m

4.9

For cadet velocity = 6 km/hr = 1.66 m/sec

Distance = vt = 1.57 × 1.66 = 2.6 m.

The cadet, 2.6 m away from tree will receive the berry on his uniform.

29. For last 6 m distance travelled s = 6 m, u = ?

t = 0.2 sec, a = g = 9.8 m/s2

S = ut  1 at2  6 = u(0.2) + 4.9 × 0.04

x m

2

 u = 5.8/0.2 = 29 m/s. For distance x, u = 0, v = 29 m/s, a = g = 9.8 m/s2

6m t=0.2 sec

S = v2  u2  292  02 = 42.05 m

2a

2  9.8

Total distance = 42.05 + 6 = 48.05 = 48 m.

30. Consider the motion of ball form A to B.

B  just above the sand (just to penetrate)

u = 0, a = 9.8 m/s2, s = 5 m

S = ut  1 at2

A

2

6 m

 5 = 0 + 1/2 (9.8)t2

 t2 = 5/4.9 = 1.02  t = 1.01.

B 10cm

 velocity at B, v = u + at = 9.8 × 1.01 (u = 0) =9.89 m/s. C

From motion of ball in sand

u1 = 9.89 m/s, v1 = 0, a = ?, s = 10 cm = 0.1 m.

a = v12  u12  0  (9.89)2 = – 490 m/s2

2s

2  0.1

The retardation in sand is 490 m/s2.

31. For elevator and coin u = 0

As the elevator descends downward with acceleration a (say)

The coin has to move more distance than 1.8 m to strike the floor. Time taken t = 1 sec.

Sc = ut  1 at2 = 0 + 1/2 g(1)2 = 1/2 g
2
Se = ut  1 at2 = u + 1/2 a(1)2 = 1/2 a
2
Total distance covered by coin is given by = 1.8 + 1/2 a = 1/2 g

a 6ft=1.8m 1/2a

 1.8 +a/2 = 9.8/2 = 4.9  a = 6.2 m/s2 = 6.2 × 3.28 = 20.34 ft/s2.

32. It is a case of projectile fired horizontally from a height.

3.8

h = 100 m, g = 9.8 m/s2 a) Time taken to reach the ground t = (2h / g)

Chapter-3

= 2 100 = 4.51 sec.
9.8
b) Horizontal range x = ut = 20 × 4.5 = 90 m.

100m

20m/s

c) Horizontal velocity remains constant through out the motion.
At A, V = 20 m/s A Vy = u + at = 0 + 9.8 × 4.5 = 44.1 m/s.

A  Vx
Vy Vr

Resultant velocity Vr = (44.1)2  202 = 48.42 m/s.

Tan  = Vy  44.1 = 2.205
Vx 20
  = tan–1 (2.205) = 60°. The ball strikes the ground with a velocity 48.42 m/s at an angle 66° with horizontal. 33. u = 40 m/s, a = g= 9.8 m/s2,  = 60° Angle of projection.

a) Maximum height h = u2 sin2   402(sin 60)2 = 60 m

2g

2  10

b) Horizontal range X = (u2 sin 2) / g = (402 sin 2(60°)) / 10 = 80 3 m.

3.9

34. g = 9.8 m/s2, 32.2 ft/s2 ; 40 yd = 120 ft horizontal range x = 120 ft, u = 64 ft/s,  = 45° We know that horizontal range X = u cos t  t = x  120 = 2.65 sec.
u cos  64 cos 45
y = u sin (t) – 1/2 gt2 = 64 1  1 (32.2)(2.65)2
2(2.65) 2

Chapter-3
10 ft 120 ft

= 7.08 ft which is less than the height of goal post.

In time 2.65, the ball travels horizontal distance 120 ft (40 yd) and vertical height 7.08 ft which is less than 10 ft. The ball will reach the goal post.

35. The goli move like a projectile.

Here h = 0.196 m

Horizontal distance X = 2 m

Acceleration g = 9.8 m/s2.

u

Time to reach the ground i.e.

0.196m

t = 2h  2  0.196 = 0.2 sec

g

9.8

2 m

Horizontal velocity with which it is projected be u.

 x = ut

 u = x  2 = 10 m/s.
t 0.2

36. Horizontal range X = 11.7 + 5 = 16.7 ft covered by te bike. g = 9.8 m/s2 = 32.2 ft/s2.

y = x tan  – gx2 sec2 
2u2
To find, minimum speed for just crossing, the ditch

y 5ft 11.7ft 5ft x

y = 0 ( A is on the x axis)

15°

15°

 x tan  = gx2 sec2   u2  gx2 sec2   gx  gx

2u2

2x tan  2 sin  cos  sin2

 u = (32.2)(16.7) (because sin 30° = 1/2)
1/ 2

 u = 32.79 ft/s = 32 ft/s. 37. tan  = 171/228   = tan–1 (171/228)

The motion of projectile (i.e. the packed) is from A. Taken reference axis at A.

  = –37° as u is below x-axis. u = 15 ft/s, g = 32.2 ft/s2, y = –171 ft

y = x tan  – x2gsec2 

2u2



 –171 = –x (0.7536) – x2g(1.568)
2(225)
 0.1125x2 + 0.7536 x – 171 = 0 x = 35.78 ft (can be calculated)

171ft

u

228ft

3.10

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SOLUTIONS TO CONCEPTS CHAPTER – 3