Question

Velocity $\left(V\right)$ versus displacement $\left(S\right)$ graph of a particle moving in a straight line is as shown in the figure. Given $V_0=4m/s$ and $S_o=20m.$ At $S=30m$, what is the acceleration of particle?

• A. $0.2m/s^2$
• B. $0.4m/s^2$
• C. $-0.2m/s^2$
• D. $-0.4m/s^2$

Answer 1

The correct option is D $-0.4m/s^2$

From $S=S_o\text{to}S=2S_o,$ velocity is given by
$V=\frac{-V_o}{S_o}S+2V_o$
$\frac{dV}{dt}=\frac{-V_o}{S_o}\left(\frac{dS}{dt}\right)$
$\text{Accleration,}a=\frac{-V_o}{S_o}(\frac{-V_o}{S_o}S+2V_o)=\frac{V_o^2}{S_o^2}.S-\frac{2V_o^2}{S_o}$
$a=\frac{V_o^2}{S_o^2}.S-\frac{2V_o^2}{S_o}$
$\text{Accleration at}S=30m$
$a={\left(\frac{4}{20}\right)}^2\times 30-\frac{2\times 4^2}{20}=-0.4m/s^2$