Rest and Motions Exercise Questions Solutions HC Verma Ch-3 Concept of Physics Vol-1 for Class-11. Step by Step Solution of Exercise Questions for Ch-3 Rest and Motions Kinematics HC Verma Concept of Physics . Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

## Rest and Motions Exercise Questions Solutions HC Verma Ch-3 Concept of Physics Vol-1 for Class-11

 Board ISC and other board Publications Bharti Bhawan Publishers Ch-3 Rest and Motions Kinematics Class 11 Vol 1st writer H C Verma Book Name Concept of Physics Topics Solution of Exercise Questions Page-Number 51, 52, 53, 54

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### HC Verma Solutions of Ch-3 Rest and Motions Exercise Questions

(Page-51)

#### Question-1

A man has to go 50 m due north, 40 m due east and 20 m due south to reach a field.

(a) What distance he has to walk to reach the field?

(b) What is his displacement from his house to the field?

(a) Distance travelled by the man = AB + BC + CD = 50 + 40 + 20 = 110 m
(b) AF = AB − BF = 50 − 20 = 30 m
Displacement = Final position − Initial position = AD #### Question-2

A particle starts from the origin, goes along the X-axis to the point (20 m, 0) and then return along the same line to the point (−20 m, 0). Find the distance and displacement of the particle during the trip.

Let the points be O(0,0), A(20 m, 0) and B(−20 m, 0).
(i) Distance travelled = OA + AB = 20 + 40 = 60 m
(ii) Displacement = OB = 20 m (in the negative direction)

#### Question-3

It is 260 km from Patna to Ranchi by air and 320 km by road. An aeroplane takes 30 minutes to go from Patna to Ranchi whereas a delux bus takes 8 hours.

(a) Find the average speed of the plane.

(b) Find the average speed of the bus.

(c) Find the average velocity of the plane.

(d) Find the average velocity of the bus.

(a) (b) (c)

The plane moves in a straight path.
Average velocity = Average speed = 520 Km / h  (Patna to Ranchi)

(d)  Straight path distance from Patna to Ranchi = Displacement of the bus = 260 km #### Question-4

When a person leaves his home for sightseeing by his car, the meter reads 12352 km. When he returns home after two hours the reading is 12416 km

(a) What is the average speed of the care during this period?

(b) What is the average of velocity?

(a) Total distance covered = 12416 − 12352 = 64 km (b) Because he returns to his house, the displacement is zero. So, the average velocity is zero

#### Question-5

An athlete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of this average acceleration.

Initial velocity, u = 0
Final velocity, v = 18 km/h = 5 m/s
Time interval, t = 2 s
Using #### Question-6

The speed of a car as a function of time is shown in the following figure. Find the distance travelled by the car in 8 seconds and its acceleration. Figure-3-E.1 #### Question-7

The acceleration of a cart started at t = 0, varies with time as shown in the following figure. Find the distance travelled in 30 seconds and draw the position-time graph Figure-3-E.2

In the first 10 seconds, At t = 10 s,
v = u + at = 0 + 5 × 10 = 50 ft/s
∴ From 10 to 20 seconds (∆t = 20 − 10 = 10 s), the particle moves with a uniform velocity of 50 ft/s.
Distance covered from t = 10 s to t = 20 s:
S2 = 50 × 10 = 500 ft

Between 20 s to 30 s, acceleration is constant, i.e., −5 ft/s2.
At 20 s, velocity is 50 ft/s.
t = 30 − 20 = 10 s Total distance travelled is 30 s:
S1 + S2 + S3
= 250 + 500 + 250
= 1000 ft
The position–time graph: #### Question-8

Figure (3 -E.3) Shows the graph of velocity versus time for a particle going along the X-axis. (a )Find the acceleration

(b) Find the distance travelled in 0 to 10s and

(c) Find the displacement in 0 to 10 s.

(a )

Slope of the v–t graph gives the acceleration.
Acceleration (b)

Area in the v–t graph gives the distance travelled.
Distance travelled = Area of ∆ABC + Area of rectangle OABD (c)

Displacement is the same as the distance travelled.
Displacement = 50 m

#### Question-9

Figure (3 -E.4) shows the graph of the x-coordinate of a particle going along the X-axis as a function of time.

(a) Find the average velocity during 0 to 10 s,

(b)  Find instantaneous velocity at 2, 5, 9 and 12s. (a)

Displacement from t = 0 s to = 10 s:
x = 100 m
Time = 10 s

Average velocity from 0 to 10 seconds, (b)

Slope of the x–t graph gives the velocity.
At 2.5 s,

slope = ⇒ vinst = 20 m/s
At 5 s, vinst = 0.
At 8 s, vinst = 20 m/s.
At 12 s, vinst = −20 m/s.

#### Question-10

From the velocity-time plot shown in figure,(3 -E.5) find the distance travelled by the particle during the first 40 seconds. Also find the average velocity during this period Area shown in the v–t graph gives the distance travelled.
∴ Distance travelled in the first 40 seconds = Area of ∆OAB + Area of ∆BCD As the displacement is zero, the average velocity is zero.

(Page-52)

#### Question-11

figure (3 -E.6)  shows x-t graph of a particle. Find the time t such that the average velocity of the particle during the period 0 to t is zero. (3 -E.6)

Consider point B1 at 12 s.

At t = 0 s, S = 20 m and at t = 12 s, S = 20 m.
For the time interval 0–12, change in the displacement is zero.

Average velocity = Displacement / Time = 0

Hence, the time is 12 seconds.

#### Question-12

A particle starts from a point A and travels along the solid curve shown in figure (3 -E.7) . Find approximately the position B of the particle such that the average velocity between the position A and B has the same direction as the instantaneous velocity at B. At position B , the instantaneous velocity of the particle has the direction along BC  #### Question-13

An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s2 for 5.0 s. Find the distance travelled during the period of acceleration.

Given:
Velocity, u = 4.0 m/s
Acceleration, = 1.2 m/s2
Time, t = 5.0 s
Distance travelled :

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