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Rest and Motions Exercise Questions Solutions HC Verma Ch-3

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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Question for Short Answer

Objective-I

Objective-II

Exercise  (Currently Open)


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?

Answer-1

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(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

HC Verma Solutions Rest and Motion Exe Ans-1

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.

Answer-2

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.

Answer-3

(a)

HC Verma Solutions Rest and Motion Exe Ans-3.1

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(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?

Answer-4

(a) Total distance covered = 12416 − 12352 = 64 km

HC Verma Solutions Rest and Motion Exe Ans-4.1

(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.

Answer-5

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

HC Verma Solutions Rest and Motion Exe Ans-5

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.

HC Verma Solutions Rest and Motion Exe Q -6

Figure-3-E.1

Answer-6

HC Verma Solutions Rest and Motion Exe Ans-6

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

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HC Verma Solutions Rest and Motion Exe Q -8

Figure-3-E.2

Answer-7

In the first 10 seconds,

HC Verma Solutions Rest and Motion Exe Ans-7.1

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:

HC Verma Solutions Rest and Motion Exe Ans-7.3

Question-8

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

HC Verma Solutions Rest and Motion Exe Q 8 figure

(a )Find the acceleration

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

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

Answer-8

(a )

Slope of the v–t graph gives the acceleration.
Acceleration

HC Verma Solutions Rest and Motion Exe Ans-8.1

(b)

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

HC Verma Solutions Rest and Motion Exe Ans-8.2

(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.

HC Verma Solutions Rest and Motion Exe Q 9

Answer-9

(a)

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

Average velocity from 0 to 10 seconds,

HC Verma Solutions Rest and Motion Exe Ans-9.1

(b)

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

slope =

HC Verma Solutions Rest and Motion Exe Ans-9.1

⇒ 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

HC Verma Solutions Rest and Motion Exe Q 10

Answer-10

Area shown in the v–t graph gives the distance travelled.
∴ Distance travelled in the first 40 seconds = Area of ∆OAB + Area of ∆BCD

HC Verma Solutions Rest and Motion Exe Ans-10

As the displacement is zero, the average velocity is zero.


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

(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.

HC Verma Solutions Rest and Motion Exe Q 11

(3 -E.6)

Answer-11

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.

HC Verma Solutions Rest and Motion Exe Q 12

Answer-12

At position B , the instantaneous velocity of the particle has the direction along BC

HC Verma Solutions Rest and Motion Exe Ans-12.2

HC Verma Solutions Rest and Motion Exe Ans-12

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.

Answer-13

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

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