Kinematics

JEE Main level, numerical answer, medium

128 of 308

Question 128 of 308

A particle is moving in a straight line such that its velocity is increasing at 5ms−15{ms}^{-1} per meter. The acceleration of the particle is _____ ms−2m{s}^{-2} at a point where its velocity is 20ms−120{ms}^{-1}.

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Answer: 100

According to the question, velocity is increasing at 5ms−15m{s}^{-1}. Therefore, dvds=5s−1\dfrac{dv}{ds}=5{s}^{-1}.
Now acceleration a=dvdt=vdvdsa=\dfrac{dv}{dt}=v\dfrac{dv}{ds}
⇒a=vdvds=20×5=100ms−2\Rightarrow a=v\dfrac{dv}{ds}=20\times 5=100m{s}^{-2}

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