Kinematics

JEE Main level, numerical answer, medium

52 of 308

Question 52 of 308

If the velocity of a body related to displacement xx is given by v=5000+24xms−1,v=\sqrt{5000+24x}m{s}^{-1}, then the acceleration of the body is _________ ms−2.m{s}^{-2}.

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Answer: 12

We know that the rate of change of linear velocity is called linear acceleration, a=dvdt=dv×dxdt×dx=vdvdxa=\dfrac{dv}{dt}=\dfrac{dv\times dx}{dt\times dx}=\dfrac{vdv}{dx}
⇒a=5000+24x×125000+24x×24\Rightarrow a=\sqrt{5000+24x}\times \dfrac{1}{2\sqrt{5000+24x}}\times 24=12ms−2=12m{s}^{-2}.

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