First slide

Instantaneous acceleration

Question

A point moves in a straight line, so that its displacement x at time t is given by $x^{2} = t^{2} + 1$ . Its acceleration is

Moderate

A

$1 / x$
B

$1 / x^{3}$
C

$- 1 / x^{2}$
D

$- 1 / x^{3}$

Solution

$x^{2} = t^{2} + 1 or 2 x \cdot \frac{d x}{d t} = 2 t or x \cdot \frac{d x}{d t} = t$
$∴$ $v = \frac{d x}{d t} = \frac{t}{x} = \frac{t}{\sqrt{t^{2} + 1}}$
$a = \frac{d v}{d t} = \frac{d^{2} x}{d t^{2}} = \frac{\sqrt{t^{2} + 1} - \frac{t^{2}}{\sqrt{t^{2} + 1}}}{(t^{2} + 1)} = \frac{1}{{(t^{2} + 1)}^{3 / 2}} = \frac{1}{x^{3}}$

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