A particle moves in a straight line so that its displacement x at any time t is given by x² = 1 + t². Its acceleration at any time t is x⁻ⁿ where n = _________. Numerical answer type. Work it out, then open the answer.

Answer: 3. Differentiate x² = 1 + t²: xv = t ⇒ v = t/x Differentiate again: v² + xa = 1 ⇒ a = (1 - v²)/(x) = (x² - t²)/(x³) x² - t² = 1, so a = x⁻³ So n = 3.

A particle moves in a straight line so that its displacement x at any… | JEE Main 2024 (6 Apr, Shift 2)

Q12JEE Main 2024 · 6 APR · SHIFT 2Medium

Velocity, Acceleration & Variable Motion

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 $n =$ _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3

Differentiate $x^2 = 1 + t^2$ : $xv = t \Rightarrow v = \dfrac{t}{x}$
Differentiate again: $v^2 + xa = 1 \Rightarrow a = \dfrac{1 - v^2}{x} = \dfrac{x^2 - t^2}{x^3}$
$x^2 - t^2 = 1$ , so $a = x^{-3}$
So $n = 3$ .

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Velocity, Acceleration & Variable Motion question from JEE Main 2024 (6 Apr, Shift 2)

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