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A particle is moving on a straight line with velocity (v) as a function of time (t) according to relation `v = (5t^(2) - 3t + 2)m//s` . Now give the answer of following questions :
Acceleration of particle at the end of `3^(rd)` second of motion is :

A

`27 m//sec^(2)`

B

`16 m//sec^(2)`

C

`39 m//sec^(2)`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
A
Acceleration of …………..
`a = (dV)/(dt) = 10t - 3`
`a at t = 3 "second is a" = 10xx3 - 3 = 27 m//s^(2)`
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