Motion in One Dimension Exe-2C Very Short Answer Physics Class-9 ICSE Selina Publishers

Motion in One Dimension Exe-2C Equations of Motion Very Short Answer Type for Class-9 ICSE Concise Physics. There is the solutions of Very short Answer and Long answer type Questions of your latest textbook which is applicable in 2023-24 academic sessionVisit official Website CISCE for detail information about ICSE Board Class-9.

Motion in One Dimension Exe-2 C Equations of Motion Very Short Answer 

(ICSE Class – 9 Physics Concise Selina Publishers)

Board ICSE
Class 9
Subject Physics
Writer / Publication Concise selina Publishers
Chapter-2 Motion in One Dimension
Exe – 2 C Equations of Motion
Topics Solution of Exe-2(C) Very Short and Long Answer Type
Academic Session 2023-2024

Exe-2C Equations of Motion Very Short Answer Type

Ch-2 Motion in One Dimension Physics Class-9 ICSE Concise

Page 60

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:

Three equations of a uniformly accelerated motion are

v = u + at

s = ut + (1/2)at2

v2 = u2 + 2as

Question 2. 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:

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 


Exe-2C Equations of Motion Long Answer Type

Motion in One Dimension Exe-2C Very Short Class-9 ICSE

Page 60

Question 1. 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:

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.

first law of motion

Acceleration = Change in velocity/Time a = (– 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 = (at)/2 s = (2att/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.

—  : End of Motion in One Dimension Exe-2C Equations of Motion Very Short and Long Answer Type  Solutions :–

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