A particle is moving in a straight line. The variation of position ' x ' as a function of time ' t ' is given as x= (t 3 -6 t 2 +20 t+15 ) m . The velocity of the body when its acceleration becomes zero is :

Physics · Motion in one Dimension

JEE Main 2024 — 29 January, Shift 2 — Question 46

A particle is moving in a straight line. The variation of position ' $x$ ' as a function of time ' $t$ ' is given as $x=\left(t^{3}-6 t^{2}+20 t+15\right) m$ . The velocity of the body when its acceleration becomes zero is :

Option A:
$4 \mathrm{~m} / \mathrm{s}$
Option B:
$8 \mathrm{~m} / \mathrm{s}$
Correct
Option C:
$10 \mathrm{~m} / \mathrm{s}$
Option D:
$6 \mathrm{~m} / \mathrm{s}$

Answer: B

Step-by-step solution

$\mathrm{x}=\mathrm{t}^{3}-6 \mathrm{t}^{2}+20 \mathrm{t}+15$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}=3 \mathrm{t}^{2}-12 \mathrm{t}+20$

$\frac{d v}{d t}=a=6 t-12$

When $\mathrm{a}=0$

$6 \mathrm{t}-12=0 ; \mathrm{t}=2 \mathrm{sec}$

At $\mathrm{t}=2 \mathrm{sec}$

$\mathrm{v}=3(2)^{2}-12(2)+20$

$\mathrm{v}=8 \mathrm{~m} / \mathrm{s}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Motion in one Dimension
Topic: Non-Uniformly Accelerated Motion