Question

The graph shows the variation of $\frac{1}{V}$ (where V is the velocity of the particle) with respect to time. Then find the value of acceleration at $t=3$ sec.

• A. $3m/s^2$
• B. $5m/s^2$
• C. $1m/s^2$
• D. $Noneofthese$

Answer 1

The correct option is A $3m/s^2$

At $t=3$

Slope=$\frac{d\left(\frac{1}{V}\right)}{dt}=-1\Rightarrow -\frac{1}{V^2}\frac{dV}{dt}=-1$

$\Rightarrow {\left(\frac{1}{\sqrt{3}}\right)}^2\frac{dV}{dt}=1\Rightarrow a=\frac{dV}{dt}=3m/s^2$