Chapter 4, Problem 1P

(a)

To determine

The total vector displacement.

Answer to Problem 1P

The displacement is $\underset{\_}{4.87\text{ km}}$ at ${28.6}^{°}$ South of West.

Explanation of Solution

Draw the figure according to given condition as.

Consider that system has three vectors.

Write the expression for first trip from point $A$ to $B$.

    ${\overrightarrow{d}}_1={\overrightarrow{v}}_1{t}_1$                                                                                                           (I)

Here, ${\overrightarrow{d}}_1$ is distance for first trip, ${\overrightarrow{v}}_1$ is speed for first trip and $t_1$ is time for first trip.

Write the expression for second trip from point $B$ to $C$ .

  ${\overrightarrow{d}}_2={\overrightarrow{v}}_2{t}_2$                                                                                                       (II)

Here, ${\overrightarrow{d}}_2$ is distance for second trip, ${\overrightarrow{v}}_2$ is speed for second trip and $t_2$ is time for second trip.

Write the expression for third trip from point $C$ to $D$.

    ${\overrightarrow{d}}_3={\overrightarrow{v}}_3{t}_3$                                                                                                       (III)

Here, ${\overrightarrow{d}}_3$ is distance for third trip, ${\overrightarrow{v}}_3$ is speed for third trip and $t_3$ is time for third trip.

Write the expression for total displacement in three trips.

    $\overrightarrow{d}={\overrightarrow{d}}_1+{\overrightarrow{d}}_2+{\overrightarrow{d}}_3$                                                                                           (IV)

Here, $\overrightarrow{d}$ is total displacement vector.

Write the expression for magnitude of the displacement.

    $\left|d\right|=\sqrt{\left(d_x{}^2\right)+\left(d_y{}^2\right)}$                                                                                      (V)

Here, $d_x$ is the x-component of distance and $d_y$ is the y-component of distance.

Write the expression for direction of vector d as.

    $\theta ={\tan }^{-1}\left(\frac{d_y}{d_x}\right)$                                                                                            (VI)

Here, $\theta$ is angle between two components.

Conclusion:

Substitute $-20.0\widehat{j}\text{ m/s}$ for ${\overrightarrow{v}}_1$ and $3.00\text{ min}$ for $t_1$ in equation (II)

    $\begin{array}{c}{\overrightarrow{d}}_1=-\left(20.0\widehat{j}\text{ m/s}\left(\text{3}\text{.00 min}\right)\right)\\ \text{=}-\left(\text{20}\text{.0 m/s}\left(\text{3}\text{.00 min}\left(\frac{\text{60 s}}{1\text{ min}}\right)\right)\right)\widehat{j}\\ =-3600\widehat{j}\text{ m}\end{array}$

Substitute $-25.0\widehat{i}\text{ m/s}$ for ${\overrightarrow{v}}_2$ and $2.00\text{ min}$ for $t_2$ in equation (III)

  $\begin{array}{c}{\overrightarrow{d}}_2=-\left(25.0\widehat{i}\text{ m/s}\left(\text{2}\text{.00 min}\right)\right)\\ \text{=}-\left(\text{25}\text{.0 m/s}\left(\text{2}\text{.00 min}\left(\frac{\text{60 s}}{1\text{ min}}\right)\right)\right)\widehat{i}\\ =-3000\widehat{i}\text{ m}\end{array}$

Substitute $\left(-30.0\cos 45°\widehat{i}+30\sin 45°\widehat{j}\right)\text{ m/s}$ for ${\overrightarrow{v}}_3$ and $1.00\text{ min}$ for $t_3$ in equation (IV)

  $\begin{array}{c}{\overrightarrow{d}}_3=\left[\left(-30.0\cos 45°\widehat{i}+30\sin 45°\widehat{j}\right)\text{ m/s}\right]\left(1.00\text{ min}\right)\\ \text{=}\left[\begin{array}{c}\left(-30.0\cos 45°\text{ m/s}\left(\text{1}\text{.00 min}\left(\frac{\text{60 s}}{1\text{ min}}\right)\right)\right)\widehat{i}+\\ \left(30.0\sin 45°\text{ m/s}\left(\text{1}\text{.00 min}\left(\frac{\text{60 s}}{1\text{ min}}\right)\right)\right)\end{array}\right]\widehat{j}\\ =\left(-1272.8\widehat{i}\text{ +}1272.8\widehat{j}\right)\text{m}\end{array}$

Substitute $-3600\widehat{j}\text{ m}$ for ${\overrightarrow{d}}_1$, $-3000\widehat{i}\text{ m}$ for ${\overrightarrow{d}}_2$ and $\left(-1272.8\widehat{i}\text{ +}1272.8\widehat{j}\right)\text{m}$ for ${\overrightarrow{d}}_3$ in equation (V).

    $\begin{array}{c}\overrightarrow{d}=\left(-3600\widehat{j}\text{ m}-3000\widehat{i}\text{ m}\right)+\left(-1272.8\widehat{i}\text{ m}+1272.8\widehat{j}\text{ m}\right)\\ =-4272.8\widehat{i}\text{ m}-2327.2\widehat{j}\text{ m}\end{array}$

Substitute $-4272.8\widehat{i}\text{ m}$ for $d_x$ and $-2327.2\widehat{j}\text{ m}$ for $d_y$ in equation (VI)

    $\begin{array}{c}d=\sqrt{{\left(-4272.8\widehat{i}\right)}^2+{\left(-2327.2\widehat{j}\right)}^2}\\ =4865.45\text{ m}\left(\frac{1\text{km}}{1000\text{ m}}\right)\\ \simeq 4.87\text{ km}\end{array}$

Substitute $-4272.8\widehat{i}\text{ m}$ for $d_x$ and $-2327.2\widehat{j}\text{ m}$ for $d_y$ in equation (VII).

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-2327.2\text{m}}{-4272.8\text{m}}\right)\\ ={\tan }^{-1}\left(0.54465\right)\\ =28.57°\\ \approx 28.6°\end{array}$

Thus the displacement is $\underset{\_}{4.87\text{ km}}$ at $28.6°$ South of West.

(b)

To determine

The average speed.

Answer to Problem 1P

The average speed is $\underset{\_}{23.3\text{ m/s}}$.

Explanation of Solution

Write the expression for average speed as.

    $v_{\text{avg}}=\frac{d_1+d_2+d_3}{t_1+t_2+t_3}$                                                                                 (VIII)

Here, $v_{\text{avg}}$ is the average speed.

Conclusion:

Substitute $3600\text{ m}$ for $d_1$, $3000\text{ m}$ for $d_2$, $1800\text{ m}$ for $d_3$, $3.00\text{ min}$ for $t_1$, $2.00\text{ min}$ for $t_2$ and $1.00\text{ min}$ for $t_3$ in equation (VIII)

    $\begin{array}{c}{v}_{\text{avg}}=\frac{\left(3600\text{ m}+3000\text{ m}+1800\text{ m}\right)}{\left(3.00\text{ min}+2.00\text{ min}+1.00\text{ min}\right)}\\ =\frac{\left(8400\text{ m}\right)}{\left(6.00\text{ min}\right)}\\ =\frac{\left(8400\text{ m}\right)}{\left(6.00\text{ min}\left(\frac{60\text{ s}}{1\text{ min}}\right)\right)}\\ =\frac{8400\text{ m}}{360\text{ s}}\end{array}$

Simplify further above equation as.

    $v_{\text{avg}}=23.3\text{ m/s}$

Thus, the average speed is $\underset{\_}{23.3\text{ m/s}}$.

(c)

To determine

The average velocity.

Answer to Problem 1P

The average velocity is $\underset{\_}{13.5\text{ m/s}}$ at $28.6°$ South of West.

Explanation of Solution

Write the expression for average velocity as.

    ${\overrightarrow{v}}_{\text{avg}}=\frac{\overrightarrow{d}}{t_1+t_2+t_3}$                                                                                           (IX)

Here, ${\overrightarrow{v}}_{\text{avg}}$ is the average velocity.

Conclusion:

Substitute $4865.45\text{ m}$ for $\overrightarrow{d}$, $t_1$, $2.00\text{ min}$ for $t_2$ and $1.00\text{ min}$ for $t_3$ in equation (IX)

    $\begin{array}{c}{\overrightarrow{v}}_{\text{avg}}=\frac{\left(4865.45\text{ m}\right)}{\left(3.00\text{ min}+2.00\text{ min}+1.00\text{ min}\right)}\\ =\frac{\left(4865.45\text{ m}\right)}{\left(6.00\text{ min}\right)}\\ =\frac{\left(4865.45\text{ m}\right)}{\left(6.00\text{ min}\left(\frac{60\text{ s}}{1\text{ min}}\right)\right)}\\ =\frac{\left(4865.45\text{ m}\right)}{\left(360\text{ s}\right)}\end{array}$

Simplify above equation as.

    ${\overrightarrow{v}}_{avg}=13.5\text{ m/s}$

Thus, the average velocity is $\underset{\_}{13.5\text{ m/s}}$ at ${28.6}^{°}$ South of West.