A particle is moving in one dimension (along x axis) under the...

A particle is moving in one dimension (along $x$ axis) under the action of a variable force. It's initial position was $16m$ right of origin. The variation of its position \left(x\right) $with time \left(t\right)$ is given as $x = – 3{t}^{3} + 18{t}^{2} + 16t,$ where $x$ is in $m and t$ is in $s .$ The velocity of the particle when its acceleration becomes zero is _________ $m{s}^{- 1} .</p>$

Correct answer

Step-by-step solution

Given:

$x = - 3{t}^{3} + 18{t}^{2} + 16t$

As we know, $v = \dfrac{dx}{dt} = - 9{t}^{2} + 36t + 16$

Now, $a = \dfrac{dv}{dt} = - 18t + 36$

For $a = 0$ , time till be $t = 2s$

Therefore, required velocity $v = - 9 ( 2{)}^{2} + 36 \times 2 + 16$

$v = 52m{s}^{- 1}$

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