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Kinematics

JEE Main level, single correct, medium

32 of 308

Question 32 of 308

The velocity of a particle is $v=\left({v}_{0}+gt+F{t}^{2}\right)m{s}^{-1}$ . Its position is $x=0$ at $t=0;$ then its displacement after time $(t=1s)$ is :

Show answer and solution

Answer: (B) ${v}_{0}+\dfrac{g}{2}+\dfrac{F}{3}$

v={v}_{0}+gt+F{t}^{2}

\dfrac{ds}{dt}={v}_{0}+gt+F{t}^{2}

\int ds={\int }_{0}^{1}\left({v}_{0}+gt+F{t}^{2}\right)dt

s={\left[{v}_{0}t+\dfrac{g{t}^{2}}{2}+\dfrac{F{t}^{3}}{3}\right]}_{0}^{1}

s={v}_{0}+\dfrac{g}{2}+\dfrac{F}{3}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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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.

The velocity of a particle is v=(v0+gt+Ft2)ms−1v=\left({v}_{0}+gt+F{t}^{2}\right)m{s}^{-1}v=(v0​+gt+Ft2)ms−1. Its position is x=0x=0x=0 at t=0;t=0;t=0; then its displacement after time (t=1s)(t=1s)(t=1s) is :

JEE Main question paper PDF

More Physics practice

The velocity of a particle is $v=\left({v}_{0}+gt+F{t}^{2}\right)m{s}^{-1}$ . Its position is $x=0$ at $t=0;$ then its displacement after time $(t=1s)$ is :