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.

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Aicharya · Accepted Answer

$\begin{aligned} & \vec{r}(t)=\left(4 t \hat{i}+2 t^2 \hat{j}+5 \hat{k}ight) \mathrm{m} \ & \frac{d \vec{r}(t)}{d t}=4 \hat{i}+\left.4 t \hat{j}ight|_{t=1} \mathrm{~ms}^{-1} \ & \vec{v}=4 \hat{i}+4 \hat{j} \mathrm{~ms}^{-1} \ & |\vec{v}|=\sqrt{4^2+4^2}=4 \sqrt{2} \mathrm{~ms}^{-1} \ & 	an \phi=\frac{v_y}{v_x}=\frac{4}{4}=1 \ & \phi=45^{\circ}\end{aligned}$

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…

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