Verified Solution • 2025–26 Session moderate

QJEEM24Phc6e3: JEE Main 2024 Physics - Class 11 JEEM24Phc6e3 Solution: A particle is moving in a straight line. The variation of…

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+20 t+15\right) m . The velocity of the body when its acceleration becomes zero is :

Multiple Choice Options

  • A 6 m/s
  • B 10 m/s
  • C 8 m/s
  • D 4 m/s

Step-by-Step Verified Solution

The position equation is given by:

x=t3−6t2+20t+15x = t^3 - 6t^2 + 20t + 15

First, compute the velocity vv by differentiating the position function with respect to time:

dxdt=v=3t2−12t+20\frac{d x}{d t} = v = 3t^2 - 12t + 20

Next, compute the acceleration aa by differentiating the velocity function with respect to time:

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

We need to find the time tt when the acceleration is zero:

6t−12=06t - 12 = 0

t=2 sect = 2 \, \mathrm{sec}

Now, find the velocity at t=2 sect = 2 \, \mathrm{sec} :

v=3(2)2−12(2)+20v = 3(2)^2 - 12(2) + 20

v=8 m/sv = 8 \, \mathrm{m/s}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring