Kinematics

JEE Main level, single correct, medium

14 of 308

Question 14 of 308

A particle is moving in a straight line. The variation of position xx as a function of time tt is given as x=(t3−6t2+20t+15)mx=\left({t}^{3}-6{t}^{2}+20t+15\right)m. The velocity of the body when its acceleration becomes zero is:

Show answer and solution

Answer: (B) 8ms−18m{s}^{-1}

Given the instantaneous position is
x=t3−6t2+20t+15...(1)x={t}^{3}-6{t}^{2}+20t+15...\left(1\right)
The velocity of the particle is given by
v=dxdt=3t2−12t+20...(2)\begin{matrix}v & = & \dfrac{dx}{dt} \\ & = & 3{t}^{2}-12t+20...\left(2\right)\end{matrix}
And, the acceleration of the particle is given by
a=dvdt=6t−12...(3)\begin{matrix}a & = & \dfrac{dv}{dt} \\ & = & 6t-12...\left(3\right)\end{matrix}
When a=0a=0, it implies
6t−12=0⇒t=2s6t-12=0 \Rightarrow t=2s
At t=2st=2s, the velocity can be calculated as follows:
v=3(2)2−12(2)+20=8ms−1\begin{matrix}v & = & 3(2{)}^{2}-12(2)+20 \\ & = & 8m{s}^{-1}\end{matrix}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Projectile motion simulationDrag the sliders and watch the numbers in this question change.Revise Kinematics & Projectile Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Kinematics (Rest and Motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.