# The position vector of a particle changes

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 $t=1$ ?

1. (1) 40

2. (2) 25

3. (3) 100

4. (4) 50

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, $\vec{a}=\frac{d \vec{v}}{d t}=30 \hat{i}-40 \hat{j}$

$\therefore a=\sqrt{30^{2}+40^{2}}=50 \mathrm{~m} / \mathrm{s}^{2}$