Kinematics

JEE Main level, numerical answer, medium

124 of 308

Question 124 of 308

A particle initially at rest starts moving from reference point x=0x=0 along xx-axis, with velocity vv that varies as v=4xms−1v=4\sqrt{x}m{s}^{-1}. The acceleration of the particle is _____ ms−2m{s}^{-2}.

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Answer: 8

Given: v=4xv=4\sqrt{x}
Instantaneous acceleration of the particle can be written as, a=vdvdxa=v\dfrac{dv}{dx}
=(4x)×(4×12x−1/2)=8ms−2=\left(4\sqrt{x}\right)\times \left(4\times \dfrac{1}{2}{x}^{-1/2}\right)=8m{s}^{-2}

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