Question

A particle is moving with acceleration $\overrightarrow{a}=(\hat{i}+\hat{j}){\text{m/s}}^2$ and its initial velocity (at $t=0$) is ${\overrightarrow{v}}_0=(\hat{i}-\hat{j})\text{m/s}$. Rate of change of speed of the particle at $t=0$ is

• A. Zero
• B. $\sqrt{2}{\text{m/s}}^2$
• C. $2\sqrt{2}{\text{m/s}}^2$
• D. $\sqrt{5}{\text{m/s}}^2$

Answer 1

The correct option is A Zero

$\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$
$\overrightarrow{v}=(\hat{i}-\hat{j})+(\hat{i}+\hat{j})t$
Speed of the particle is
$\left|\overrightarrow{v}\right|=\sqrt{(1+t{)}^2+(t-1{)}^2}$

Rate of change of speed is
$\frac{d\left|\overrightarrow{v}\right|}{dt}=\frac{2(1+t)+2(t-1)}{2\sqrt{(1+t{)}^2+(t-1{)}^2}}$
At $t=0$
$\frac{d\left|\overrightarrow{v}\right|}{dt}=0$