Chapter 4, Problem 77PQ

(a)

To determine

Find the velocity of the particle as a function of time.

Answer to Problem 77PQ

The velocity of the particle as a function of time is $\left(2.00+\frac{2\sqrt{3}}{3}{t}^{3/2}\right)\widehat{i}+\left(7.00-\frac{t^2}{2}\right)\widehat{j}$.

Explanation of Solution

Write the equation for velocity.

    $\begin{array}{l}{\displaystyle {\int }_i^{f}d\overrightarrow{v}}={\displaystyle {\int }_i^f\overrightarrow{a}dt}=\Delta \overrightarrow{v}\\ {\displaystyle {\int }_i^f\overrightarrow{a}dt}=\overrightarrow{v}-v_i\end{array}$                                                             (I)

Here, $\overrightarrow{a}$ is the time in acceleration, $\overrightarrow{v}$ is the velocity and $v_i$ is the initial velocity.

Conclusion:

Substitute $\left(2.00\widehat{i}+7.00\widehat{j}\right)\text{ m/s}$ for $v_i$ and $\left(\sqrt{3}{t}^{1/2}\widehat{i}-t\widehat{j}\right){\text{ m/s}}^2$ for $\overrightarrow{a}$ in equation I.

    $\begin{array}{l}{\displaystyle {\int }_0^t\left(\left(\sqrt{3}{t}^{1/2}\widehat{i}-t\widehat{j}\right){\text{ m/s}}^2\right)dt}=\overrightarrow{v}-\left(\left(2.00\widehat{i}+7.00\widehat{j}\right)\text{ m/s}\right)\\ \overrightarrow{v}=\left(2.00\widehat{i}+7.00\widehat{j}\right)\text{ m/s}+{\displaystyle {\int }_0^t\sqrt{3}{t}^{1/2}\widehat{i}dt}-{\displaystyle {\int }_0^{t}t\widehat{j}dt}\\ =\left(2.00\widehat{i}+7.00\widehat{j}\right)\text{ m/s}+{\left[\sqrt{3}\frac{t^{3/2}}{3/2}\right]}_0^t\widehat{i}-{\left[\frac{t^2}{2}\right]}_0^t\widehat{j}\\ =\left(2.00+\frac{2\sqrt{3}}{3}{t}^{3/2}\right)\widehat{i}+\left(7.00-\frac{t^2}{2}\right)\widehat{j}\end{array}$

Therefore, the velocity of the particle as a function of time is $\left(2.00+\frac{2\sqrt{3}}{3}{t}^{3/2}\right)\widehat{i}+\left(7.00-\frac{t^2}{2}\right)\widehat{j}$.

(b)

To determine

Find the position of the particle as a function of time.

Answer to Problem 77PQ

The position of the particle as a function of time is $\left(2.00t+\frac{4\sqrt{3}}{15}{t}^{5/2}\right)\widehat{i}+\left(7.00t-\frac{t^3}{6}\right)\widehat{j}$.

Explanation of Solution

Write the equation for position.

    $\begin{array}{l}{\displaystyle {\int }_i^{f}d\overrightarrow{r}}={\displaystyle {\int }_i^f\overrightarrow{v}dt}=\Delta \overrightarrow{r}\\ {\displaystyle {\int }_i^f\overrightarrow{v}dt}=\overrightarrow{r}-r_i\end{array}$                                                             (II)

Here, $r_i$ is the initial position.

Conclusion:

Substitute $0$ for $r_i$ and $\left(2.00+\frac{2\sqrt{3}}{3}{t}^{3/2}\right)\widehat{i}+\left(7.00-\frac{t^2}{2}\right)\widehat{j}$ for $\overrightarrow{v}$ in equation II.

    $\begin{array}{l}{\displaystyle {\int }_i^f\left(\left(2.00+\frac{2\sqrt{3}}{3}{t}^{3/2}\right)\widehat{i}+\left(7.00-\frac{t^2}{2}\right)\widehat{j}\right)dt}=\overrightarrow{r}-0\\ \overrightarrow{r}={\displaystyle {\int }_0^t\left(2.00+\frac{2\sqrt{3}}{3}{t}^{3/2}\right)\widehat{i}dt}-{\displaystyle {\int }_0^t\left(7.00-\frac{t^2}{2}\right)\widehat{j}dt}\\ =\left(2.00t+\frac{4\sqrt{3}}{15}{t}^{5/2}\right)\widehat{i}+\left(7.00t-\frac{t^3}{6}\right)\widehat{j}\end{array}$

Therefore, the position of the particle as a function of time is $\left(2.00t+\frac{4\sqrt{3}}{15}{t}^{5/2}\right)\widehat{i}+\left(7.00t-\frac{t^3}{6}\right)\widehat{j}$.