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Kinematics

JEE Main level, single correct, medium

54 of 308

Question 54 of 308

A particle of mass $M$ originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation $F={F}_{0}\left[1-{\left(\dfrac{t-T}{T}\right)}^{2}\right]$ where ${F}_{0}$ and $T$ are constants. The force acts only for the time interval $2T$ . The velocity $v$ of the particle after time $2T$ is:

Show answer and solution

Answer: (C) $\dfrac{4{F}_{0}T}{3M}$

t=0,u=0

a=\dfrac{{F}_{o}}{M}-\dfrac{{F}_{o}}{M{T}^{2}}(t-T{)}^{2}=\dfrac{dv}{dt}

{\int }_{0}^{v}dv={\int }_{t=0}^{2T}\left(\dfrac{{F}_{o}}{M}-\dfrac{{F}_{o}}{M{T}^{2}}(t-T{)}^{2}\right)dt

V={\left[\dfrac{{F}_{o}}{M}t\right]}_{o}^{2T}-\dfrac{{F}_{o}}{M{T}^{2}}{\left[\dfrac{{t}^{3}}{3}-{t}^{2}T+{T}^{2}t\right]}_{0}^{2T}

V=\dfrac{4{F}_{o}T}{3M}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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