Question

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

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

Answer 1

The correct option is D $4\sqrt{3}m{s}^{-2}$

We know slope of v-t graph is

$\frac{\Delta v}{\Delta t}$ i.e., acceleration

so at

we know that acceleration $a=\frac{vdv}{ds}$

$\frac{dv}{ds}=$ slope of v-s graph

at $s=3$m the slope is $m=-\tan {60}^o$

so slope $=m=\frac{dv}{ds}=-\sqrt{3}{s}^{-1}$

$\Rightarrow$ acceleration $=v\times (-\sqrt{3})=4\frac{m}{s}(-\sqrt{3})s^{-1}$

$a=-4\sqrt{3}m/s^2$

Value $=|a|=4\sqrt{3}m/s^2$