Question

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 1

Correct Answer: 3

\[x^2 = 1 + t^2\]
Differentiating with respect to \(t\):
\[2x \frac{dx}{dt} = 2t\]
\[x \cdot v = t \quad \text{(where \(v = \frac{dx}{dt}\))}\]
Differentiating again:
\[x \frac{dv}{dt} + v \frac{dx}{dt} = 1\]
\[x \cdot a + v^2 = 1 \quad \text{(where \(a = \frac{dv}{dt}\))}\]
Simplify:
\[a = \frac{1 - v^2}{x} = \frac{1 - t^2 / x^2}{x}\]
\[a = \frac{1}{x^3} = x^{-3}\]