Q105 / 200 · NEET 2023 (Manipur) · Physics
The position of a particle is given by
$\vec{r}(t)=4 t \hat{i}+2 t^2 \hat{j}+5 \hat{k}$
where $t$ is in seconds and $r$ in metre. Find the magnitude and direction of velocity $v(t)$, at $t=1 \mathrm{~s}$, with respect to $x$-axis.
- A$3 \sqrt{2} \mathrm{~ms}^{-1}, 30^{\circ}$
- B$3 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}$
- Correct answerC$4 \sqrt{2} \mathrm{~ms}^{-1}, 45^{\circ}$
- D$4 \sqrt{2} \mathrm{~ms}^{-1}, 60^{\circ}$
Solution
$\begin{aligned} & \vec{r}(t)=\left(4 t \hat{i}+2 t^2 \hat{j}+5 \hat{k}\right) \mathrm{m} \\ & \frac{d \vec{r}(t)}{d t}=4 \hat{i}+\left.4 t \hat{j}\right|_{t=1} \mathrm{~ms}^{-1} \\ & \vec{v}=4 \hat{i}+4 \hat{j} \mathrm{~ms}^{-1} \\ & |\vec{v}|=\sqrt{4^2+4^2}=4 \sqrt{2} \mathrm{~ms}^{-1} \\ & \tan \phi=\frac{v_y}{v_x}=\frac{4}{4}=1 \\ & \phi=45^{\circ}\end{aligned}$
Hence, the correct answer is option: 'C'.