Kinematics

JEE Main level, single correct, medium

134 of 308

Question 134 of 308

The position of a particle as a function of time t,t, is given by x(t)=at+bt2−ct3x\left(t\right)=at+b{t}^{2}-c{t}^{3} where a,ba,b and cc are constants. When the particles zero acceleration, then its velocity will be:

Show answer and solution

Answer: (A) a+b23ca+\dfrac{{b}^{2}}{3c}

x(t)=at+bt2−ct3x(t)=at+b{t}^{2}-c{t}^{3}
velocity v(t)=ddt(x)=ddt(at+bt2−ct3)v(t)=\dfrac{d}{dt}(x)=\dfrac{d}{dt}\left(at+b{t}^{2}-c{t}^{3}\right)
=a+2bt−3ct2=a+2bt-3c{t}^{2}
Acceleration =ddtv(t)=2b−6tc=\dfrac{d}{dt}v(t)=2b-6tc
acceleration=0⇒2b−6tc=0\text{acceleration}=0 \Rightarrow 2b-6tc=0
t=b3ct=\dfrac{b}{3c}
∴∴ velocity when t=b3c,t=\dfrac{b}{3c},
v(t=b3c)=a+2b(b3c)−3(b3c)2{v}_{\left(t=\dfrac{b}{3c}\right)}=a+2b\left(\dfrac{b}{3c}\right)-3{\left(\dfrac{b}{3c}\right)}^{2}
=a+2b23c−b23c=a+\dfrac{2{b}^{2}}{3c}-\dfrac{{b}^{2}}{3c}
=a+b23c=a+\dfrac{{b}^{2}}{3c}

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 (Non-uniform Motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.