CLASS - IX Physics (Chapter - MOTION)
larks questions
An object is moving up an inclined plan. Its velocity changes from 15m/s to 10m/s in two
seconds. What is its acceleration? (ans. -2.5m/s)
A body covered a distance of x metre along a semicircular path. Calculate the magnitude of
displacement of the body, and the ratio of distance to displacement?
A particle moving with an initial velocity of 5m/s is subjected to a uniform acceleration of -
2.5m/s2. Find the displacement in the next 4 sec. (ans. 0)
Answers
- Initial velocity of object (u) = 15m/s
Final velocity of object (v) = 10 m/s
Time taken (t) = 2 seconds
As, we know formula for acceleration is -
(v-u)/t
So, Acceleration = (10-15)/2
= -5/2
= -2.5 m/s²
Answer) Acceleration of body is -2.5 m/s²
2.Let the radius of the semi-circular path be r
So, the distance covered by the path is πr=x m⇒ r=x/π
And the displacement = 2r=2Xx/π
So the ratio of distance to displacement = x:2x/π=π:2x
3.Given:
Initial velocity =u=5m/s
Acceleration=a=-2.5m/s2
Time =t=4sec
Displacement =s =?
From Second equation of motion:
S=ut+ 1/2 at^2
=5×4+1/2*(-2.5)4*4
=20-25*16/2*10
=20-20
=0 m
please make sure me brainlist and don't forget to give thanks
a=v-u/t so 10-15/2 =-5/2=-2.5 ms-2 ques have said on inclined plane that's why it opposes motion and it is in negative sign
s= ut +1/2at^2
s=5*4+1/2x2.5*4.5)^2
this will give u the answer