The position vector of a moving body at any instant of time is given as $\vec{r}=\left(5 t^{2} \hat{

Question

The position vector of a moving body at any instant of time is given as $\vec{r}=\left(5 t^{2} \hat{i}-5 t \hat{j}ight) \mathrm{m}$. The magnitude and direction of velocity at $t=2 \mathrm{~s}$ is,\

Professor PC Thomas Classes · Answer

(d) : Given, position vector\ $(\vec{r})=\left(5 t^{2} \hat{i}-5 t \hat{j}ight) \mathrm{m}$\ $\vec{v}=\frac{d \vec{r}}{d t}=(10 t \hat{i}-5 \hat{j}) \mathrm{m} / \mathrm{s}$\ At, $t=2 \mathrm{sec}$\ $\vec{v}=20 \hat{i}-5 \hat{j}$\ From the graph,\ $	an 	heta=\frac{20}{5}=4$\ $	heta=	an ^{-1}(4)$\ Magnitude,\ $=\sqrt{(20)^{2}+(5)^{2}}=\sqrt{425}=5 \sqrt{17} \mathrm{~m} / \mathrm{s}$\ Hence, the magnitude is $5 \sqrt{17} \mathrm{~m} / \mathrm{s}$ making an angle of $	an ^{-1}(4)$ with negative $Y$-axis.\