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Kinematics

JEE Main level, single correct, medium

295 of 308

Question 295 of 308

A ball is spun with angular acceleration $\alpha =6{t}^{2}-2t$ where $t$ is in second and $\alpha$ is in $rad{s}^{-2}$ . At $t=0$ , the ball has angular velocity of $10rad{s}^{-1}$ and angular position of $4rad$ . The most appropriate expression for the angular position of the ball is

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Answer: (B) $\dfrac{{t}^{4}}{2}-\dfrac{{t}^{3}}{3}+10t+4$

Given:

\alpha =6{t}^{2}-2t

Using relation

\alpha =\dfrac{d\omega }{dt}=6{t}^{2}-2t

Integrating the above,

{\int }_{10}^{\omega }d\omega ={\int }_{0}^{t}\left(6{t}^{2}-2t\right)dt

\Rightarrow \omega -10=2{t}^{3}-{t}^{2}

Now,

\omega =\dfrac{d\theta }{dt}=10+2{t}^{3}-{t}^{2}

{\int }_{4}^{\theta }d\theta ={\int }_{0}^{t}\left(10+2{t}^{3}-{t}^{2}\right)dt

Integrating the above relation,

\theta -4=10t+\dfrac{{t}^{4}}{2}-\dfrac{{t}^{3}}{3}

Thus, angular position of the ball is

\theta =\dfrac{{t}^{4}}{2}-\dfrac{{t}^{3}}{3}+10t+4

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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