Kinematics

JEE Main level, single correct, medium

30 of 308

Question 30 of 308

The relation between time tt and distance xx for a moving body is given as t=mx2+nxt=m{x}^{2}+nx, where mm and nn are constants. The retardation of the motion is: (When vv stands for velocity)

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Answer: (A) 2mv32m{v}^{3}

t=mx2+nxt=m{x}^{2}+nx
1v=dtdx=2mx+n\dfrac{1}{v}=\dfrac{\text{d}t}{\text{d}x}=2mx+n
V=12mx+nV=\dfrac{1}{2mx+n}
dvdt=−2m(2mx+n)2(dxdt)\dfrac{\text{d}v}{\text{d}t}=-\dfrac{2m}{(2mx+n{)}^{2}}\left(\dfrac{\text{d}x}{\text{d}t}\right)
a=−(2m)v3a=-(2m){v}^{3}

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