A particle moves in a straight line so that its displacement x at any time t is given by $\mathrm{x}^{2}=1+\mathrm{t}^{2}$. Its acceleration at any time t is $\mathrm{x}^{-\mathrm{n}}$ where $\mathrm{n}=$ ______.

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Kinematics

A particle moves in a straight line so that its displacement x at any time t is given by

x^{2} = 1 + t^{2}

. Its acceleration at any time t is

x^{- n}

where

n =

______.

Answer: 3

Solution

x^{2} = 1 + t^{2} \Rightarrow 2 x \overset{x}{˙} = 2 t \Rightarrow x \overset{x}{˙} = t

. Differentiating again:

\overset{x}{˙}^{2} + x \overset{x}{¨} = 1

, so

\overset{x}{¨} = \frac{1 - x ˙ ^{2}}{x}

. Since

\overset{x}{˙} = t / x

\overset{x}{˙}^{2} = t^{2} / x^{2} = (x^{2} - 1) / x^{2}

, so

1 - \overset{x}{˙}^{2} = 1/ x^{2}

. Hence

\overset{x}{¨} = \frac{1}{x ^{3}} = x^{- 3}

, so

n = 3

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Integer answer

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