First slide

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Question

velocity v and the distance traveled by a particle 'x' has the relation $2 + 3 v^{2} = x^{2}$ then its acceleration is

Easy

A

x
B

$\frac{x}{2}$
C

$\frac{x}{3}$
D

$\frac{x}{4}$

Solution

$2 + 3 v^{2} = x^{2}$
0 + 6 v . a = 2x . v

$a = \frac{x}{3}$

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