Solved Examples

1. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds ?

Sol. Distance covered in one round = 40 m

Time taken = 40s

Total time of journey = 2 min. 20s = 140 s

Total distance covered = = 140 M

Total rounds = = 3.5 round

After every complete round, farmer will be at the starting point, hence after 3.5 rounds farmer will be at the diagonal ends of the field.

Displacement = =

2. In a long distance race the athletes were expected to take four rounds of the track such that the line of finish was same as the line of start. Suppose the length of the track was 200 m

(i) What is the total distance to be covered by the athletes?

(ii) What is the displacement of the athletes when they touch the finish line?

(iii) Is the motion of the athletes uniform or non-uniform.

(iv) Is the displacement of an athlete and the distance moved by him at the end of the race equal?

Sol.

(i) Total distance covered = 4 × 200m = 800 m

(ii) As the athletes finish at the starting line

Displacement = final position-initial positive = r_A_r_A = 0

(iii) Motion is non-uniform as the direction of motion of the athlete is changing while running on the track.

(iv) Displacement and distance moved are not equal.

 

3. On a 120km track, a train travels the first 30km with a uniform speed of 30km/h. How fast must the train travel the next 90km so as to average 60km/h for the entire trip ?

Sol. Total distance d = 120km

Average speed V_av = 60km/h

Total time = t = ?

Average speed =

V_av =

t = Putting the values

t = = 2h .............(i)

Distance travelled in first part of trip

d₁ = 30km

Speed in first part of the trip v₁ = 30km/h

Time taken in first part of trip

t₁ = ?

t₁ = putting the values

t₁ = = 1h

Time taken left to complete second part of the trip

t₂ = t _ t₁ = 2 _ 1 = 1h

Distance to be covered in second part of the trip

d₂ = 90 km

required speed in second part

v₂ = ?

speed =

v₂ = = = 90 km/h

 

4. Nisha swims in a 90 m long pool. She covers 180m in one minute by swimming from one end to the other and back along the same straigh path. Find the average speed and average velocity of Nisha.

Sol. Total distance = 180m

Total displacement = 0

Time taken t = 1min. = 60s

Average speed (V_av) =

V_av = = 3 m/s

Average velocity (V_av) =

V_av = = 0 m/s

 

5. A bus decreases its speed from 80 km/h to 60 km/h is 5 second. Find the acceleration of the bus.

Sol. u = 80 km/h = 80 × = 22.22 m/s

v = 60 km/h = 60 × = 16.67 m/s

time t = 5s

Acceleration (a) =

a = = _ 1.11 m/s²

 

6. A train starting from a railway station and moving with uniform acceleration, attains a speed 40 km/h in 10 minutes. Find its acceleration.

Sol. u = 0 (starting from rest)

v = 40 km/h = 40 × = 11.11 m/s

time t = 10 minutes = 600s

Acceleration (a) =

a = = 0.0185 m/s²

 

7. A bus between Kota to Jaipur passed the 100 km, 160 km and 220 km points at 10.30 am, 11.30 am and 1.30 pm. Find the average speed of the bus during each of the following intervals :

(a) 10.30 am to 11.30 am, (b) 11.30 am to 1.30 pm and (c) 10.30 am to 1.30 pm.

Sol. (a) The distance covered between 10.30 am and 11.30 am is 160 km _ 100 km = 60 km. The time interval is 1 hour. The average speed during this interval is _

v₁ = = 60 km/h

(b) The distance covered between 11.30 am and 1.30 pm is 220 km _ 160 = 60 km. The time interval is 2 hours. The average speed during this interval is

v₂ = = 30 km/h

(c) The distance covered between 10.30 am and 1.30 pm is 220 km _ 100 km = 120 km. The time interval is 3 hours. The average speed during this interval is _

v₃ = = 40 km/h

 

8. The average speed of a bicycle, an athlete and car are 18 km/h, 7 m/s and 2 km/min. respectively. Which of the three is the fastest and which is the slowest ?

Sol. 18km/h = = = 5 m/s

2 km/min = == 33.3 m/s

Thus the average speeds of the bicycle, the athlete and the car are 5 m/s, 7 m/s and 33.3 m/s respectively. So the car is the fastest and the bicycle is the slowest.

 

9. An object is sliding down on inclined plane. The velocity changes at a constant rate from 10 cm/s to 15 cm/s in 2 seconds. What is its acceleration?

Sol. The situation is shown in figure. Let us take BA as the positive direction. The velocity at t = 0 is u = 10 cm/s and that at t = 25 is v = 15 cm/s.

Thus, a = = == 2.5 cm/s²

The acceleration is positive, which means it is in the direction BA.

 

10. Figure shows distance-time graph of two objects A and B which object is moving with greater speed when both are moving?

Sol. The line for object B makes a longer angle with the time-axis. Its slope is longer than the slope of the line for object A. Thus, the speed of B is greater than that of A.

 

11. Figure represents the speed time graph for a particle. Find the distance covered by the particle between t = 10 min. and t = 30 min.

Sol. We draw perpendicular lines from the 10-minute point and the 30-minute point to the time-axis (fig.) The distance covered is equal to the area of the rectangle ABCD its value is

ABCD = (30 min._10 min.) × (10 km/h)

= 20 min. × 10 km/h

= h × 10 km/h = km.

 

12. Find the distance coverd by a particle during the time interval t = 0 to t = 20 s for which the speed-time graph is shown in figure.

Sol. The distance covered in the time interval 0 to 20 s. is equal to the area of the shaded triangle. It is

× base × height.

= × (20 s) × (20 m/s) = 200 m.

 

13. Figure shows the velocity-time graphs for two objects, A and B, moving along the same direction. Which object has greater acceleration ?

Sol. The slope of the velocity-time graph of B is greater than that for A, Thus, the acceleration of B is greater than that of A.

 

14. A particle moves through a distance of 3 m due east and then 4m due north.

(a) How much is the net distance traversed ? (b) What is the magnitude of the net displacement?

Sol. The situation shown in figure. The particle starts from O. It moves through a distance of 3m due east to reach A and then through a distance of 4m due north to reach B.

(a) The total distance moved is 3m 4m = 7 m,

(b) The magnitude of the net displacement is OB. In the right_angled triangle OAB,

OB² = OA² AB² = (3m)² (4m)² = 9m² 16m² = 25m² OR OB = 5 m,

 

15. A car moves 30 km. in 30 min. and the next 30 km. in 40 min. calculate the average speed for the entire journey.

Sol. Given, the total time taken is 30 min. 40 min. = 70 min. and the total distance traversed is _

30 km 30 km = 60 km. The average speed is

v_av = km/hr = 51.4 km/h

 

16. A boy runs for 10 min. at a uniform speed of 9 km/h. At what speed should he run for the next 20 min. So that the average speed comes to 12 km/h?

Sol. Total time = 10 min 20 min = 30 min.

The average speed is 12km/h. using s = vt, the distance covered in 30 min is

12 km/h × 30min = = 6km

The distance covered in the first 10 min is _

9 km/h × 10 min. = = 1.5 km

Thus, he has to cover 6 km _ 1.5 km = 4.5 km. in the next 20 min. The speed required is

= 13.5 km/h

 

17. A particle was at rest from 9:00 am to 9:30 am. It moved at a uniform speed of 10 km/h from 9:30 am to 10:00 am. Find the average speed between (a) 9:00 am and 10:00 am (b) 9:15 am and 10:00 am

Sol. (a) The distance moved by the particle between 9:30 am and 10:00 am is

s = vt = = 5 km.

This is also the distance moved between 9:00 am and 10:00 am. Thus, the average speed during this interval is

v_av = = 5 km/h

(b) The distance moved between 9:30 am and 10:00 am is 5km. This is also the distance moved in the interval 9:15 am to 10:00 am. The average speed during this interval is _

v_av = = km/h 6.67 km/h.

 

18. A boy is running on a straight road. He runs 500 m towards north in 2.10 minutes and then turns back and runs 200 m in 1.00 minute. Calculate :-

(a) His average speed and magnitude of average velocity during first 2.10 minutes, and

(b) His average speed and magnitude of average velocity during the whole journey.

Sol. (a) Total distance = 500 m

Total time = 2.10 minutes = 130 s

Magnitude of displacement = 500 m

Average speed = = 3.85 ms^_1

Magnitude of average velocity = = 3.85 ms^_1

Note : This example shows that the average speed = average velocity if the motion is in one direction.

(b) Total distance = 500 m 200 m = 700 m

Total time = 2.10 1.00 = 3.10 minutes = 190 s

Magnitude of total displacement = 500 m _ 200 m = 300 m

Average speed = = 3.68 ms^_1

Magnitude of average velocity = = 1.58 ms^_1

Note : This example shows that average speed is greater than the magnitude of average velocity if the direction of motion changes.

 

19. It is estimated that the radio signal takes 1.27 seconds to reach the earth from the surface of the moon. Calculate the distance of the moon from the earth. Speed of radio signal = 3 × 10⁸ ms^_1 (speed of light in air).

Sol. Here, time = 1.27 s

speed = 3 × 10⁸ ms^_1

distance = ?

Using distance = speed × time, we get

distance = 3 × 10⁸ ms^_1 × 1.27 s = 3.81 × 10⁸ m = 3.81 = 10⁵ km

 

20. A wireless signal is sent to earth from a spacecraft. This signal reaches the earth in 300 seconds. Calculate the distance of the spacecraft from the earth. Given, speed of the signal = 3 × 10⁸ ms^_1.

Sol. Here, t = 300 s

Speed, v = 3 × 10⁸ ms^_1

Distance = ?

Using distance = speed × time, we get

distance = 3 × 10⁸ ms^_1 × 300 s = 9 × 10¹⁰ m = 9 = 10⁷ km

Thus, distance of spacecraft from earth = 9 × 10⁷ km.

 

21. A sound is heard 5 seconds later than the lightning is seen in the sky on a rainy day. Find the distance of the location of lightning. Given speed of sound = 346 ms^_1.

Sol. Here, t = 5 s

Speed, v = 346 ms^_1

Distance = ?

Using distance = speed × time, we get

distance = 346 ms^_1 × 5 s = 1730 m

Thus, distance of the location of lightning = 1730 m