Question

A particle moves on circular path of radius 5m with constant speed 5m/s. Find the magnitude of its average acceleration when it completes half revolution

Answer 1

Answer:

10/π m/s²

Explanation:

Given data:

let radius=r

r=5 m

v=5 m/s

we should know that since the body is moving with constant velocity it will experience only centripetal acceleration and tangential acceleration will be zero.

When the body completes half revolution then its velocity vectors will be in mutually opposite directions.

let initial velocity be v

vi=v n

(where n is the unit direction vector)

final velocity is

vf=v (-n)

We have the formula

Aav=Δv/Δt

(where Aav=average acceleration)

Δv=vf-vi

=v (-n) - v (n)

=-2v (n)

|Δv|=2v

Δt=πr/v

Aav=2v/πr/v

=2v²/πr

Plugging in the values

=2*5*5/π*5

=10/π m/s²

Hope this helps.