Question

Let position of a particle be expressed as $\overrightarrow{r}=(t^3\hat{i}+t^2\hat{j})m$, (where $t$ is time in second.). Velocity at t = 2 sec.

• A. $2m/s$
• B. $12.6m/s$
• C. $4m/s$
• D. $Zero$

Answer 1

The correct option is B $12.6m/s$

Given that,

Time $t=2s$

Position vector $\overrightarrow{r}=(t^3\hat{i}+t^2\hat{j})m$

Now, on differentiate of position vector

$\frac{d\overrightarrow{r}}{dt}=(3t^2\hat{i}+2t\hat{j})$

Now, at $t=2s$

The velocity is

$\overrightarrow{v}=12\hat{i}+4\hat{j}$

Now, the magnitude of velocity

$v=\sqrt{{\left(12\right)}^2+{\left(4\right)}^2}$

$v=\sqrt{160}$

$v=12.6m/s$

Hence, the velocity is $12.6m/s$