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**Solved Examples****Ques 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?****Ans:** 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, the farmer will be at the starting point, hence after 3.5 rounds farmer will be at the diagonal ends of the field.

Displacement = = **Ques 2 : **

**Ans:**

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

**Ans:**Total distance d = 120 km

Average speed V_{av} = 60 km/h

Total time = t = ?

Average speed =

V_{av} =

t = Putting the values

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

Distance travelled in first part of trip

d_{1} = 30 km

Speed in first part of the trip v_{1} = 30 km/h

Time taken in first part of trip

t_{1} = ?

t_{1} = putting the values

t_{1} = = 1 h

Time taken left to complete second part of the trip

t_{2} = t - t_{1} = 2 - 1 = 1h

Distance to be covered in second part of the trip

d_{2} = 90 km

reQues uired speed in second part

v_{2} = ?

speed =

v_{2} = = = 90 km/h **Ques 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 straight path. Find the average speed and average velocity of Nisha. ****Ans:**Total distance = 180 m

Total displacement = 0

Time taken t = 1min. = 60 s

Average speed (V_{av}) =

V_{av} = = 3 m/s|

Average velocity

(V_{av}) =

V_{av} = ** **= 0 m/s**Ques ****5: ****A bus decreases its speed from 80 km/h to 60 km/h is 5 seconds. Find the acceleration of the bus.****Ans: **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/s^{2}**Ques 6 : **

v = 40 km/h = 40 × = 11.11 m/s

time t = 10 minutes = 600s

Acceleration (a) =

a = = 0.0185 m/s^{2}**Ques ****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.

(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 v

(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 _v

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.

Thus, a = = == 2.5 cm/s

The acceleration is positive, which means it is in the direction BA.**Ques 10 : Figure shows the distance-time graph of two objects A and B which object is moving with greater speed when both are moving? **

**Ans: **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.**Ques 11 : **

**Ans: **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.**Ques 12 : **

**Ans: **The distance covered in the time interval 0 to 20 s. is eQues ual to the area of the shaded triangle. It is

× base × height.

= × (20 s) × (20 m/s) = 200 m.**Ques 13 : **

**Ans: **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.**Ques 14 : **

(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,

OB^{2} = OA^{2} AB^{2} = (3m)^{2} (4m)^{2} = 9m^{2} 16m^{2} = 25m^{2} **OR** OB = 5 m, **Ques 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.**

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

v_{av} = km/hr = 51.4 km/h **Ques 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?**

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

12 km/h × 30 min = = 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**Ques 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**

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

(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.**Ques 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 :-**

Total time = 2.10 minutes = 130 s

Magnitude of displacement = 500 m

Average speed = = 3.85 ms

**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.**Ques 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 × 10^{8} ms^{-1} (speed of light in air).**

speed = 3 × 10

distance = ?

Using distance = speed × time, we get

distance = 3 × 10

**Ques 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 × 10^{8} ms^{-1}.**

Speed, v = 3 × 10

Distance = ?

Using distance = speed × time, we get

distance = 3 × 10

Thus, distance of spacecraft from earth = 9 × 10

Speed, v = 346 ms

Distance = ?

Using distance = speed × time, we get

distance = 346 ms

Thus, distance of the location of lightning = 1730 m

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