Question

A particle starts from the origin with velocity $\mathbf{5}\hat{\mathbf{i}}\mathbf{ }\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$at role="math" localid="1663603319140" $\mathbf{t}\mathbf{=}\mathbf{0}$and moves in the$\mathbf{xy}$plane with a varying acceleration given by$\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{\left(}\mathbf{6}\sqrt{\mathbf{t}}\hat{\mathbf{j}}\mathbf{\right)}$, where$\overrightarrow{\mathbf{a}}$is in meters per second squared and$\mathit{t}$is in seconds.

(a) Determine the velocity of the particle as a function of time.

(b) Determine the position of the particle as a function of time.

Short answer

(a) The particle's velocity is$\left(5\hat{\mathrm{i}}+4t^{\frac{3}{2}}\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$.

(b) The position of the particle as a function of time is $\left(5t\hat{i}+1.6t^{\frac{5}{2}}\hat{\mathrm{j}}\right)\mathrm{m}$.

Step-by-step solution

Acceleration formula:

Write the expression for the acceleration

$\overrightarrow{\mathbf{a}}\mathbf{=}\frac{\mathbf{d}\overrightarrow{\mathbf{v}}}{\mathbf{d}\mathbf{t}}$

Here,$\overrightarrow{\mathbf{v}}$is the instantaneous velocity of the particle,$\mathbf{t}$is the time and$\overrightarrow{\mathbf{a}}$is the acceleration of the particle.

(a) Determine the velocity of the particle as a function of time:

The initial velocity of the particle at the origin is $5\hat{i}\mathrm{m}/\mathrm{s}$at $t=0$and its acceleration when it moves in the $xy$plane is $\overrightarrow{a}=\left(6\sqrt{t}\hat{\mathrm{j}}\right)$.

Rearrange the above equation for and integrate

$\begin{array}{c}{\int }_0^{t}d\overrightarrow{v}={\int }_0^t\overrightarrow{a}dt\\ \overrightarrow{v}\left(t\right)-\overrightarrow{v}\left(0\right)={\int }_0^t\overrightarrow{a}dt\\ \overrightarrow{v}\left(t\right)=\overrightarrow{v}\left(0\right)+{\int }_0^t\overrightarrow{a}dt\end{array}$

Substitute $5\hat{\mathrm{i}}\mathrm{m}/\mathrm{s}$for $\overrightarrow{v}\left(0\right)$and $\left(6\sqrt{t } j\right)\mathrm{m}/{\mathrm{s}}^2$for $\overrightarrow{a}$to find the$\overrightarrow{v}\left(t\right)$

$\begin{array}{c}\overrightarrow{v}\left(t\right)=5\widehat{\mathrm{i} }\mathrm{m}/\mathrm{s}+{\int }_0^t\left(6\sqrt{t}\hat{\mathrm{j}} \mathrm{m}/{\mathrm{s}}^2\right) tdt\\ =5\hat{\mathrm{i}} \mathrm{m}/\mathrm{s}+{\left[\frac{6t^{\frac{3}{2}}}{3/2}\hat{\mathrm{j}} \mathrm{m}/\mathrm{s}\right]}_0^t\\ =\left(5\hat{\mathrm{i}}+4t^{\frac{3}{2}}\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}\end{array}$

Therefore, the velocity of the particle as a function of time is $\left(5\hat{\mathrm{i}}+4t^{\frac{3}{2}}\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$.

(b) The position of the particle as a function of time:

Write the relation between the velocity and position

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

Here, $\overrightarrow{r}$is the position vector.

Rearrange the above equation for $\overrightarrow{v}$and integrate

role="math" localid="1663604043785" $\begin{array}{c}{\int }_0^{t}d\overrightarrow{r}={\int }_0^t\overrightarrow{v}dt\\ \overrightarrow{r}\left(t\right)-\overrightarrow{r}\left(0\right)={\int }_0^t\overrightarrow{v}dt\\ \overrightarrow{r}\left(t\right)=\overrightarrow{r}\left(0\right)+{\int }_0^t\overrightarrow{v}dt\end{array}$

Substitute $5\hat{\mathrm{i}}\mathrm{m}/\mathrm{s}$$\overrightarrow{v}\left(0\right),0$for $\overrightarrow{r}$, and $\left(5\hat{\mathrm{i}}+4t^{\frac{3}{2}}\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$for $\overrightarrow{r}$to find the $s$in terms of $t$.

$\begin{array}{c}\overrightarrow{r}\left(t\right)=0\mathrm{m}+{\int }_0^t\left(5\hat{\mathrm{i}}+4t^{\frac{3}{2}}\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}dt\\ ={\left[5t\hat{\mathrm{i}}+\frac{4t^{\frac{5}{2}}}{5/2}\hat{\mathrm{j}}\right]}_0^t\mathrm{m}\\ =\left(5t\hat{\mathrm{i}}+1.6t^{\frac{5}{2}}\hat{\mathrm{j}}\right)\mathrm{m}\end{array}$

Therefore, the position of the particle as a function of time is $\left(5t\hat{i}+1.6t^{\frac{5}{2}}\hat{\mathrm{j}}\right)\mathrm{m}$.