Question

Given that for a two-dimensional motion, velocity $\overrightarrow{v}=t\hat{i}+t^2\hat{j}$. If particle starts from $(0,0)$ then find out the magnitude of displacement in next $2\text{seconds}$.

• A. $\frac{10}{3}\text{m}$
• B. $\frac{5}{3}\text{m}$
• C. $\frac{7}{9}\text{m}$
• D. $\frac{8}{3}\text{m}$

Answer 1

The correct option is A $\frac{10}{3}\text{m}$

Given,

$\overrightarrow{v}=t\hat{i}+t^2\hat{j}$

Comparing the above equation with, $\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$, we get

$v_x=t$ and $v_y=t^2$

$v_x=\frac{dx}{dt}=t$

Now, integrating the above equation between $t=0$ to $t=2$.

${\int }_0^{x}dx={\int }_0^{2}tdt$

$\therefore x=2\text{m}$

Now,
$v_y=\frac{dy}{dt}=t^2$

Again, integrating the above equation between $t=0$ to $t=2$

${\int }_0^{y}dy={\int }_0^2{t}^{2}dt$

$\therefore y=\frac{2^3}{3}=\frac{8}{3}\text{m}$

So, net displacement after $2\text{sec}$.

$d_{net}=\sqrt{x^2+y^2}$

$d_{net}=\sqrt{4+\frac{64}{9}}=\sqrt{\frac{100}{9}}$

$\therefore d_{net}=\frac{10}{3}\text{m}$

Hence, option $\left(\text{A}\right)$ is the correct answer.