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