Exercise-2C Motion in One Dimension Concise ICSE Class-9 Selina

Exercise-2C Motion in One Dimension Concise ICSE Class-9 Selina Publishers Chapter-2 .Step By Step ICSE Selina Concise Solutions of Chapter-2 Motion in One Dimension  with Exercise-2(A), Exercise-2(B) and Exercise-2(C) including Numerical and MCQ Questions Solved . Visit official Website CISCE for detail information about ICSE Board Class-9.

Board	ICSE
Publications	Selina Publication
Subject	Physics
Class	9th
Chapter-2	Motion in One Dimension  Exe-2(c)
Book Name	Concise
Topics	Solution of Exercise-2(A),  MCQ-2(A),  Numericals-2(A),    Exercise-2(B),  MCQ-2(B),  Numericals-2(B),   Exercise-2(C), MCQ 2(C) ,  Numericals-2(C),
Academic Session	2021-2022

Exercise-2C Motion in One Dimension Concise ICSE Class-9 Selina Publishers Chapter-2

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Select Topics 

 Exercise-2(B),  MCQ-2(B),  Numericals-2(B),

Note :-  Before Viewing Selina Concise Physics Solutions of Chapter-2 Motion in One Dimension Class-9 . Read the whole chapter carefully and Solved all example of Exercise-2 Motion in One DimensionClass-9. 

Focus on Chapter-2 Motion in One Dimension in Scalar and vector quantities, distance, speed, velocity acceleration.  Motion in One Dimension in graph of distance – time and speed- time.  Equations of uniformly accelerated motion with derivations in Motion in One Dimension .

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EXERCISE-2(C) Selina Concise Physics Solutions for ICSE Class-9 Chapter-2 Motion in One Dimension 

Page 54

Question 1

Write three equations of uniformly accelerated motion relating the initial velocity (u), final velocity (v), time (t), acceleration (a) and displacement (S).

Answer 1

Three equations of a uniformly accelerated motion are

v = u + at

s = ut + (1/2)at²

v² = u² + 2as

Question 2

Derive the following equations for uniformly accelerated motion:

(i) v = u + at

(ii) s = ut + (1/2) at²

(iii) v² = u² + 2aS

where the symbols have their usual meanings.

Answer 2 Motion in One Dimension Concise

Derivation of equations of motion

First equation of motion:

Consider a particle moving along a straight line with uniform acceleration ‘a’. At t = 0, let the particle be at A and u be its initial velocity, and at t = t, let v be its final velocity.

Acceleration = Change in velocity/Time a = (v – u)/t at = v – uv = u+ at … First equation of motion.

Second equation of motion:

Average velocity = Total distance traveled/Total time takenAverage velocity = s/t …(1)

Average velocity can be written as (u+v)/2 Average velocity = (u+v)/2 …(2)

From equations (1) and (2) s/t = (u+v)/2 …(3)

The first equation of motion is v = u + at.

Substituting the value of v in equation (3), we get

s/t = (u + u + at)/2 s = (2u + at) t/2 = 2ut + at²/2 = 2ut/2 + at²/2

s = ut + (1/2) at² …Second equation of motion.

Third equation of motion: 

The first equation of motion is v = u + at. v – u = at … (1)

Average velocity = s/t …(2)

Average velocity =(u+v)/2 …(3)

From equation (2) and equation (3) we get,

(u + v)/2 = s/t …(4)

Multiplying eq (1) and eq (4) we get,

(v – u)(v + u) = at × (2s/t) (v – u)(v + u) = 2as

[We make the use of the identity a² – b² = (a + b) (a – b)]

v² – u² = 2as …Third equation of motion.

Question 3

Write an expression for the distance S covered in time t by a body which is initially at rest and starts moving with a constant acceleration a.

Answer 3

Distance = s, time = t, initial velocity u = 0 and acceleration = a.

Using the second equation of motion and substituting the above values we get,

s = ut + (1/2) at² 

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MULTIPLE CHOICE TYPE-2(C) Selina Concise Physics Solutions for ICSE Class-9 Chapter-2 Motion in One Dimension 

Page 54

Question 1

The correct equation of motion is

(a) v = u + As

(b) v = ut + a

(c) s = ut + (1/2)

(d) v = u + at

Answer 1

v = u + at so option (a) is correct

Question 2

A car starting from rest accelerates uniformly to acquire a speed 20 km h^-1 in 30 min. The distance travelled by a car in this time interval will be

(a) 600 km

(b) 5 km

(c) 6 km

(d) 10 km

Answer 2

5 km

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NUMERICAL-2(C) Selina Concise Physics Solutions for ICSE Class-9 Chapter-2 Motion in One Dimension 

Page 54

Question 1

A body starts from rest with uniform acceleration 2 m s^-2. Find the distance covered by the body in 2 s.

Answer 1

Initial velocity u = 0

Acceleration a = 2 m/s²

Time t = 2 s

Let ‘S’ be the distance covered.

Using the second equation of motion,

S = ut + (1/2) at²

S = 0 + (1/2) (2) (2) ²

S = 4 m

Question 2  (Exercise-2C Motion in One Dimension Concise ICSE)

A body starts with an initial velocity of 10m s^-1 and acceleration 5 m s^-2. Find the distance covered by it in 5 s.

Answer 2

Initial velocity u = 10 m/s

Acceleration a = 5 m/s²

Time t = 5s

Let ‘S’ be the distance covered.

Using the second equation of motion,

S = ut + (1/2) at²

S = (10)(5) + (1/2) (5) (5) ²

S = 50 + 62.5

S = 112.5 m

Question 3 Motion in One Dimension Concise

A vehicle is accelerating on a straight road. Its velocity at any instant is 30 km/h. After 2s, it is 33.6 km/h, and after further 2 s, it is 37.2 km/h. Find the acceleration of the vehicle in m s^-2. Is the acceleration uniform?

Answer 3

Acceleration = Change in velocity/time taken

In the first two seconds,

Acceleration = [(33.6 – 30)/2] km/h²

= 1.8 km/h²

= 0.5 m/s² …(i)

In the next two seconds,

Acceleration = [(37.2 – 33.6)/2] km/h²

= 1.8 km/h²

= 0.5 m/s^–2…(ii)

From (i) and (ii), we can say that the acceleration is uniform.

Question 4

A body, initially at rest, starts moving with a constant acceleration 2 m s^-2. Calculate: (i) the velocity acquired and (ii) the distance travelled in 5 s.

Answer 4

Initial velocity u = 0 m/s

Acceleration a = 2 m/s²

Time t = 5 s

(i) Let ‘v’ be the final velocity.

Then, (v – u)/5 = 2

v = 10 m/s^-1

(ii) Let ‘s’ be the distance travelled.

Using the third equation of motion,

v² – u² = 2as

We get,

(10) ² – (0) ² = 2(2) (s)

Thus, s = (100/4) m = 25 m

Question 5

A bullet initially moving with a velocity 20 m/s strikes a target and comes to rest after penetrating  a  distance 10  cm into the target. Calculate the retardation caused by the target.

Answer 5

Initial velocity u = 20 m/s

Final velocity v = 0

Distance travelled s = 10 cm = 0.1 m

Let acceleration be ‘a’.

Using the third equation of motion,

v² – u² = 2as

We get,

(0) ² – (20) ² = 2(a) (0.1)

a = -(400/0.2) m/s²

a = -2000 m/s²

Thus, retardation = 2000 m/s^-2

Question 6 (Exercise-2C Motion in One Dimension Concise ICSE)

A train moving with a velocity of 20 m s^-1 is brought to rest by applying brakes in 5 s. Calculate the retardation.