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 $t=3\text{sec}$

• A. $3{\text{m/s}}^2$
• B. $5{\text{m/s}}^2$
• C. $1{\text{m/s}}^2$
• D. $0.5{\text{m/s}}^2$

Answer 1

The correct option is A $3{\text{m/s}}^2$

Slope $=\frac{d\left(\frac{1}{v}\right)}{dt}=-1$
$\Rightarrow -\frac{1}{v^2}\frac{dv}{dt}=-1$
$\Rightarrow {\left(\frac{1}{\sqrt{3}}\right)}^2\frac{dv}{dt}=1$
$\Rightarrow a=\frac{dv}{dt}=3{\text{m/s}}^2$