Question:
The position vector of a particle changes with time according to the relation $\vec{r}(\mathrm{t})=15 \mathrm{t}^{2} \hat{i}+\left(4-20 \mathrm{t}^{2}\right) \hat{j}$. What
is the magnitude of the acceleration at $t=1$ ?
Correct Option: , 4
Solution:
(4) $\vec{r}=15 t^{2} \hat{i}+\left(4-20 t^{2}\right) \hat{j}$
$\vec{v}=\frac{d \vec{r}}{d t}=30 t \hat{i}-40 \hat{t j}$
Acceleration, $\vec{a}=\frac{d \vec{v}}{d t}=30 \hat{i}-40 \hat{j}$
$\therefore a=\sqrt{30^{2}+40^{2}}=50 \mathrm{~m} / \mathrm{s}^{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.