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Q.

A particle is moving along a straight line whose velocity-displacement graph is as shown in the figure. What is the magnitude of acceleration when displacement is $3 m$ ?

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a

$3 \sqrt{3} {ms}^{- 2}$

b

$- 4 \sqrt{3} {ms}^{- 2}$

c

$\sqrt{3} {ms}^{- 2}$

d

$\frac{4}{\sqrt{3}} {ms}^{- 2}$

answer is A.

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Detailed Solution

$a = v \cdot \frac{d v}{d s} = (4) (- \tan 60 °) = - 4 \sqrt{3} m / s^{2}$

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