MCQ: Equations of Motion - Class 9 MCQ

Velocity acquired by the two bodies falling from rest through a distance his given as v = 

Given that :

Initial velocity, u = 0.2 m/s. 

Final velocity, v = 0 

Acceleration, a = 0.02 m/s^2.

v = u + at 

0 = 0.2 - 0.02 x t

0.02 x t = 0.2 

t = 0.2 / 0.02 = 10 sec.

Thus, the body will take 10 seconds to stop.

The third equation of motion is :- 
 ( v² - u² = 2as) 
Here,
u = initial velocity 
v =  final velocity 
a =  acceleration
s =  distance

Consider a velocity - time graph for a uniformly accelerated body starting from rest is represented as follows.

u = velocity at time t1 
v =  velocity at time t2 
If acceleration is represented as a, then, acceleration is defined as the rate of change in velocity. 
⇒  a = v-u / t2 - t1 
⇒ a = v-u /t 
Or,v-u = at 
⇒ v= u + at 

Given:Acceleration = 6 m/s^2Time = 10 secAs car start from rest so intial velocity will be zero.u = 0S = ut + 1/2(at^2)S = 0 + 1/2 x 6 x 100S = 300 m.