Question

The velocity of a particle moving in a curvilinear path varies with time as $\overrightarrow{v}=2t^2\hat{i}+t^3\hat{j}m/s$ where $t$ is in seconds. At the end of $t=1s$, the value of tangential acceleration is

• A. $\frac{11}{\sqrt{10}}m/s^2$
• B. $\frac{11}{\sqrt{5}}m/s^2$
• C. $\sqrt{5}m/s^2$
• D. $11\sqrt{5}m/s^2$

Answer 1

The correct option is B $\frac{11}{\sqrt{5}}m/s^2$

Given $\overrightarrow{v}=2t^2\hat{i}+t^3\hat{j}$
$\overrightarrow{a}=\frac{dv}{dt}=4t\hat{i}+3t^2\hat{j}$
At $t=1s$ : $\overrightarrow{v}=2\hat{i}+\hat{j}$
$|\overrightarrow{v}|=\sqrt{4+1}=\sqrt{5}m/s$
$\overrightarrow{a}=4\hat{i}+3\hat{j}$
$|\overrightarrow{a}|=\sqrt{16+9}=\sqrt{25}m/s^2=5m/s^2$
and $\overrightarrow{a}.\overrightarrow{v}=(4\hat{i}+3\hat{j}).(2\hat{i}+\hat{j})$
$=8+3=11$
Tangential acceleration
$a_t=a\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{v}}{|\overrightarrow{v}|}=\frac{11}{\sqrt{5}}m/s^2$