Question

Velocity of a particle at any time t is $\overrightarrow{v}=(2\hat{i}+2t\hat{j})$ $m/s$. If acceleration and displacement of particle at $t=1sec$ be $a$ and $s$ respectively, then

• A. $s=(2\hat{i}+\hat{j})$ m
• B. we can use $\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$
• C. $a=\left(2\hat{j}\right)m/s^2$
• D. None of the above are correct

Answer 1

Answer: (A), (B), (C)

The correct options are
A $s=(2\hat{i}+\hat{j})$ m
B we can use $\overrightarrow{v}=\overrightarrow{u}+\overrightarrow{a}t$
C $a=\left(2\hat{j}\right)m/s^2$

$\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=0\hat{i}+2\hat{j}=2\hat{j}m/s^2$
Since acceleration is constant, equations of motion can be used, thus B is correct. Also, C is correct.

Now, $\frac{d\overrightarrow{s}}{dt}=\overrightarrow{v}=2\hat{i}+2t\hat{j}$
Integrate to get, $\overrightarrow{s}=(2t+c_1)\hat{i}+(t^2+c_2)\hat{j}$

For total distance, take initial position as ($0,0$).

Thus at $t=1$, $\overrightarrow{s}=(2\hat{i}+\hat{j})m$