A particle is moving in one dimension (along x axis) under the action of a variable force. Its 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. Numerical answer type. Work it out, then open the answer.

Answer: 52. v = dx/dt = -9t² + 36t + 16 a = dv/dt = -18t + 36 = 0 ⇒ t = 2 s v(2) = -36 + 72 + 16 = 52 So the velocity is 52 m/s.

A particle is moving in one dimension (along x axis) under the action… | JEE Main 2024 (1 Feb, Shift 1)

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Velocity, Acceleration & Variable Motion

A particle is moving in one dimension (along $x$ axis) under the action of a variable force. Its initial position was $16\text{ m}$ right of origin. The variation of its position $(x)$ with time $(t)$ is given as $x = -3t^3 + 18t^2 + 16t$ , where $x$ is in $\text{m}$ and $t$ is in $\text{s}$ .

The velocity of the particle when its acceleration becomes zero is _________ $\text{m/s}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 52

$v = \dfrac{dx}{dt} = -9t^2 + 36t + 16$
$a = \dfrac{dv}{dt} = -18t + 36 = 0 \Rightarrow t = 2\text{ s}$
$v(2) = -36 + 72 + 16 = 52$
So the velocity is $52\text{ m/s}$ .

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