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

$x^2 = 1 + t^2$
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:
$a = \frac{1 - v^2}{x} = \frac{1 - t^2 / x^2}{x}$
$a = \frac{1}{x^3} = x^{-3}$