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A car moves around a curve at a constant...

A car moves around a curve at a constant speed. When the car goes around the arc subtending `60^(@)` at the centre, then the ratio of magnitude of instantaneous acceleration to average acceleration over the `60^(@)` arc is `:`

A

`(pi)/(3)`

B

`(pi)/(6)`

C

`(2pi)/(3)`

D

`(5pi)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
A

`|Delta V| = sqrt(v^(2)+v^(2)-2v^(2) cos 60^(@))`
`= va_(av) = (|Delta vec(v)|)/(Delta t)=(v)/(t) =(3v^(2))/(pi R)`
`rArr a=(v^(2))/(R") , (a_(i))/(a_(av)) = (v^(2) pi R)/(R xx 3v^(2)) = (pi)/(3)`
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