Question

A particle is moving along a straight line whose velocity displacement curve is as shown in figure

What will be the acceleration of particle when it's displacement is 3m?

• A. $2\sqrt{3}$ $ms^{-2}$
• B. $4\sqrt{3}$ $ms^{-2}$
• C. $-2\sqrt{3}$ $ms^{-2}$
• D. $\sqrt{3}$ $ms^{-2}$

Answer 1

Answer: (C)

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

Answer 2

The acceleration a is related to velocity and displacement by the formula:

$a = V \frac{dV}{dS}$

From the graph, at S = 3m, V = 2 m/s.

The slope of the V-S curve at S = 3m, which is $\frac{dV}{dS}$, is $\tan(120^\circ) = -\sqrt{3}$.

Therefore, $a = (2) \times (-\sqrt{3}) = -2\sqrt{3}$ $ms^{-2}$.