Question: The position of a particle is given by
$$ \vec{r}(t)=4 t \hat{i}+2 t^2 \hat{j}+5 \hat{k} $$ where $\mathrm{t}$ is in seconds and $\mathrm{r}$ in metre. Find the magnitude and direction of velocity $v(t)$, at $t=1 \mathrm{~s}$, with respect to $\mathrm{x}$-axis
A) $4 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}$
B) $4 \sqrt{2} \mathrm{~ms}^{-1}, 60^{\circ}$
C) $3 \sqrt{2} \mathrm{~ms}^{-1}, 30^{\circ}$
D) $3 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}$
Answer: $4 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}$
Sol:
$\begin{aligned} & \overrightarrow{\mathrm{V}}=\frac{\overrightarrow{\mathrm{dr}}}{\mathrm{dt}}=4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+0 \hat{\mathrm{k}} \ & \text { at } \mathrm{t}=1 \mathrm{sec} \ & \overrightarrow{\mathrm{V}}=4 \hat{\mathrm{i}}+4(1) \hat{\mathrm{j}} \ & |\overrightarrow{\mathrm{V}}|=\sqrt{4^2+4^2}=4 \sqrt{2} \ & \tan \alpha=\frac{4}{4}=1 \ & \alpha=45^{\circ} \end{aligned}$
