First slide

Instantaneous acceleration

Question

A particle starting from rest and moving along a straight line has acceleration (a) given as a function of displacement (x) by the equation $a = bx$ , where b is a positive constant. The velocity of the particle when it has a displacement d is

Moderate

A

$\sqrt{b} d$
B

$\sqrt{2 b} d$
C

$\sqrt{b / 2} d$
D

$(\sqrt{b} d) / 2$

Solution

$a = bx \Rightarrow v d v = bxdx \Rightarrow \int_{0}^{v} v d v = \int_{0}^{d} bxdx$
From this $v^{2} / 2 = {bd}^{2} / 2$ , hence v = $\sqrt{b} d$

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