Chapter 3, Problem 50P

(a)

To determine

The change in velocity during the trip.

Answer to Problem 50P

The change in velocity during the trip is $\underset{\_}{\Delta \overrightarrow{v}=176\text{km/h at 24}°\text{south of the east}}$.

Explanation of Solution

Consider the figure 1 showing initial and final velocities

Let $+x$ be the east direction, and $+y$ be he north direction.

Write the expression for change in velocity

$\Delta \overrightarrow{v}=v_{\text{f}}-v_{\text{i}}$ (I)

Here, $\Delta \overrightarrow{v}$ is the change in velocity, $v_{\text{f}}$ is the final velocity, and $v_{\text{i}}$ is the initial velocity

Write the expression for magnitude of change in velocity

$\left|\Delta \overrightarrow{v}\right|=\sqrt{{\left(\Delta {\text{v}}_x\right)}^2+{\left(\Delta {\text{v}}_y\right)}^2}$ (II)

Write the expression for direction of change in velocity

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\text{opposite}}{\text{adjacent}}\right)\\ ={\tan }^{-1}\left(\frac{{\text{v}}_y}{{\text{v}}_x}\right)\end{array}$ (III)

Conclusion:

Substitute $100\text{km/h}$ for $v_{\text{f}}$, and $90\text{km/h}$ for $v_{\text{i}}$ in equation (I)

$\Delta \overrightarrow{v}=100\text{km/h}\left(\text{SE}\right)-90\text{km/h}\left(\text{W}\right)$ (IV)

If west changes to east, equation (IV) become

$\Delta \overrightarrow{v}=100\text{km/h}\left(\text{SE}\right)+90\text{km/h}\left(\text{E}\right)$ (V)

The $x$ and $y$ component of velocity are

${\text{v}}_x=\left(100\text{km/h}\right)\cos 315°+90\text{km/h}$ (VI)

${\text{v}}_y=\left(100\text{km/h}\right)\sin 315°$ (VII)

Substitute equation (VI) and (VII) in (II)

$\begin{array}{c}\left|\Delta \overrightarrow{v}\right|=\sqrt{{\left(\left(100\text{km/h}\right)\cos 315°+90\text{km/h}\right)}^2+{\left(\left(100\text{km/h}\right)\sin 315°\right)}^2}\\ =176\text{km/h}\end{array}$

Substitute equation (VI) and (VII) in (III)

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(100\text{km/h}\right)\sin 315°}{\left(100\text{km/h}\right)\cos 315°+90\text{km/h}}\right)\\ =-24°\\ =24°\text{south of east}\end{array}$

Therefore, the change in velocity during the trip is $\underset{\_}{\Delta \overrightarrow{v}=176\text{km/h at 24}°\text{south of the east}}$.

(b)

To determine

The average acceleration during the trip.

Answer to Problem 50P

The average acceleration during the trip $\underset{\_}{{\overrightarrow{a}}_{\text{av}}=284{\text{km/h}}^2\text{at 24}°\text{south of east}}$.

Explanation of Solution

Write the expression of change in time for the three distances

$\Delta t=\frac{d_1}{{\text{v}}_1}+\frac{d_2}{{\text{v}}_2}+\frac{d_3}{{\text{v}}_3}$ (I)

Here, $d_1$ is the distance travelled in west direction, ${\text{v}}_1$ is the velocity in west direction, $d_2$ is the distance travelled in south direction, ${\text{v}}_2$ is the velocity in south direction, $d_3$ is the distance travelled in south east direction, and ${\text{v}}_3$ is the velocity in south east direction

Write expression for average velocity

${\overrightarrow{a}}_{av}=\frac{\overrightarrow{\Delta v}}{\Delta t}$ (II)

Here $\overrightarrow{a_{av}}$ is the average velocity, $\Delta \overrightarrow{\text{v}}$ is the change in velocity, $\Delta t$ is the change in time

Conclusion:

Substitute $16\text{km}$ for $d1$, $8\text{km}$ for $d_2$, $34\text{km}$ for $d_3$, $90\text{km/h}$ for $v_1$, $80\text{km/h}$ for ${\text{v}}_2$, and $100\text{km/h}$ in equation (I)

$\begin{array}{c}\Delta t=\frac{16\text{km}}{90\text{km/h}}+\frac{8\text{km}}{80\text{km/h}}+\frac{34\text{km}}{100\text{km/h}}\\ =0.62\text{h}\end{array}$

Substitute $176\text{km/h}\text{at 24}°\text{south of east}$ for $\Delta \overrightarrow{\text{v}}$, and $0.62\text{h}$ in equation (II)

$\begin{array}{c}{\overrightarrow{a}}_{av}=\frac{176\text{km/h}\text{at 24}°\text{south of east}}{0.62\text{h}}\\ =284{\text{km/h}}^2\end{array}$

Therefore, the average acceleration during the trip $\underset{\_}{{\overrightarrow{a}}_{\text{av}}=284{\text{km/h}}^2\text{at 24}°\text{south of east}}$.