A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial…

A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position x with time t is given as x=3t3+18t2+16t, where x is in m and t is in s. The velocity of the particle when its acceleration becomes zero is _________ m s-1.

Solution

Given:

x=-3t3+18t2+16t

As we know, v=dxdt=-9t2+36t+16

Now, a=dvdt=-18t+36

For a=0, time till be t=2 s

Therefore, required velocity v=-9(2)2+36×2+16

v=52 m s-1

Asked in: JEE Main 2024 (01 Feb Shift 1)

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