# 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

**Select Topics **

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

**Exercise-2(C),**

**MCQ-2(C)**

**,**

**Numericals-2(C),**

** **

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 Dimension**Class-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 .**

**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)at^{2}

v^{2} = u^{2} + 2as

**Question 2**

Derive the following equations for uniformly accelerated motion:

**(i) v = u + at**

**(ii) s = ut + (1/2) at ^{2}**

**(iii) v ^{2} = u^{2} + 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 – u**v = u+ at … First equation of motion.**

**Second equation of motion:**

Average velocity = *Total* *distance* *traveled/Total* *time* *taken*Average 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 = (2*u *+ *at*) *t/*2 = 2*ut *+ *at*^{2}/2 = 2*ut/*2 + *at*^{2}/2

**s = ut + (1/2) at ^{2} …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 × (2*s/t)* (v – u)(v + u) = 2as

[We make the use of the identity a^{2} – b^{2} = (a + b) (a – b)]

**v ^{2} – u^{2} = 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) at ^{2} **

**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

**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/s^{2}

Time t = 2 s

Let ‘S’ be the distance covered.

Using the second equation of motion,

S = ut + (1/2) at^{2}

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/s^{2}

Time t = 5s

Let ‘S’ be the distance covered.

Using the second equation of motion,

S = ut + (1/2) at^{2}

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/h^{2}

= 1.8 km/h^{2}

= 0.5 m/s^{2} …(i)

In the next two seconds,

Acceleration = [(37.2 – 33.6)/2] km/h^{2}

= 1.8 km/h^{2}

= 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/s^{2}

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,

v^{2 }– u^{2 }= 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,

v^{2 }– u^{2 }= 2as

We get,

(0)^{ 2} – (20)^{ 2} = 2(a) (0.1)

a = -(400/0.2) m/s^{2}

a = -2000 m/s^{2}

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.

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