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Q.

A particle moves along X-axis. The velocity of the particle as function of $x$ is $V^{2} = 12 x - 3 x^{3}$ (in SI). Find velocity of particle at the instant when acceleration is $1 {.5 m/s}^{2}$ .

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a

Zero

b

4 m/s

c

3 m/s

d

1.5 m/s

answer is A.

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Detailed Solution

$V^{2} = 12 x - 3 x^{3}$

$2 V \frac{d V}{d x} = 12 - 9 x^{2}$

$a = 6 - \frac{9}{2} x^{2}$

$\frac{3}{2} = 6 - \frac{9}{2} x^{2} \to x = 1 m$

$V = \sqrt{(12 X 1) - 3 X 1^{3}} \to V = 3 m / s$

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