Question

The position vector of a particle changes with time according to the relation $\overrightarrow{r}\left(t\right)=15t^2\hat{i}+(4-20t^2)\hat{j}$. What is the magnitude of the acceleration at $t=1$?

• A. $40$
• B. $100$
• C. $50$
• D. $25$

Answer 1

The correct option is C $50$

The position vector of particle is given as,

$\overrightarrow{r}\left(t\right)=15t^2\hat{i}+(4-20t^2)\hat{j}$

Velocity of particle is,

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\frac{d}{dt}[15t^2\hat{i}+(4-20t^2\left)\hat{j}\right]$
$=30t\hat{i}-40t\hat{j}$
$\text{Acceleration,}\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=30\hat{i}-40\hat{j}$

$\therefore a_{t=1}=\sqrt{{30}^2+(-40{)}^2}=50$

Hence, $\left(C\right)$ is the correct answer.