Physics · Motion in one Dimension

JEE Main 2024 — 1 February, Shift 1 — Question 56

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 $(\mathrm{t})$ is given as $x=-3 t^{3}+18 t^{2}+16 t$ , where $x$ is in $m$ and $t$ is in $s$ . The velocity of the particle when its acceleration becomes zero is $\_\_\_\_\_\_\_$ $\mathrm{m} / \mathrm{s}$ .

Answer: 52

Numerical answer — enter this value.

Step-by-step solution

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

$\begin{array}{*{35}{r}}{} & v=-9{{t}^{2}}+36+16 \\{} & a=-18t+36 \\{} & a=0\text{ }\!\!~\!\!\text{ at }\!\!~\!\!\text{ }t=2s \\{} & v=-9{{(2)}^{2}}+36\times 2+16 \\{} & v=52\text{ }\!\!~\!\!\text{ m}/\text{s} \\\end{array}$

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Exam: JEE Main 2024
Paper: 1 February, Shift 1
Subject: Physics
Chapter: Motion in one Dimension
Topic: Non-Uniformly Accelerated Motion