Question

The diagram shows variation of $\frac{1}{v}$ with respect to time (where $v$ is velocity in $\text{m/s}$).

What is the instantaneous acceleration of the body at $t=3\text{sec}$. $\left(\text{in}\frac{m}{s^2}\right)$.

• A. $2$
• B. $\frac{1}{3}$
• C. $1$
• D. None of these

Answer 1

The correct option is B $\frac{1}{3}$

From graph,
The slope of the curve at ($3,\sqrt{3}$) is $-\tan {45}^{\circ }$
Hence,
$-\tan {45}^{\circ }=\frac{d\left(\frac{1}{v}\right)}{dt}=-v^{-2}\frac{dv}{dt}$
$\frac{dv}{dt}=v^2$
At $t=3,\frac{1}{v}=\sqrt{3}$
$\frac{dv}{dt}=\frac{1}{3}m/{\text{sec}}^2$