Kinematics

JEE Main level, numerical answer, medium

8 of 308

Question 8 of 308

A particle moves in a straight line so that its displacement xx at any time tt is given by x2=1+t2x^2=1+t^2. Its acceleration at any time t\mathrm{t} is x−nx^{-\mathrm{n}} where n=\mathrm{n}= ___________

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Answer: 3

x2=1+t22xdxdt=2txv=txdvdt+vdxdt=1x⋅a+v2=1a=1−v2x=1−t2/x2xa=1x3=x−3\begin{aligned} & \mathrm{x}^2=1+\mathrm{t}^2 \\ & 2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t} \\ & \mathrm{xv}=\mathrm{t} \\ & \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v} \frac{\mathrm{dx}}{\mathrm{dt}}=1 \\ & \mathrm{x} \cdot \mathrm{a}+\mathrm{v}^2=1 \\ & \mathrm{a}=\frac{1-\mathrm{v}^2}{\mathrm{x}}=\frac{1-\mathrm{t}^2 / \mathrm{x}^2}{\mathrm{x}} \\ & \mathrm{a}=\frac{1}{\mathrm{x}^3}=\mathrm{x}^{-3}\end{aligned}

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