First slide

Instantaneous acceleration

Question

The distance $(x)$ travelled by a particle in time, t, is given by $t = 2 x^{2} + 3 x .$ If ‘u’ is the velocity, then acceleration will be:

Easy

A

$- 4 u^{3}$
B

$- 3 u^{4}$
C

$4 u^{2}$
D

$- 3 u^{3}$

Solution

Differentiating with respect to t,
$1 = 4 x \frac{d x}{d t} + 3 \frac{d x}{d t} v = \frac{1}{4 x + 3}$
differentiating again:

$0 = 4 v^{2} + 4 x \frac{d^{2} x}{d t^{2}} + 3 \frac{d^{2} x}{d t^{2}} a (4 x + 3) = - 4 v^{2} a = - 4 v^{3}$

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