Chapter 3, Problem 42P

(a)

To determine

The average speed of the trip.

Answer to Problem 42P

The average speed of the trip is $102\text{km/h}$.

Explanation of Solution

The speed of the speed boat for first $20.0\min$ is $108\text{km/h}$. The speed of the boat for next $10.0\min$ is $90.0\text{km/h}$.

Write the formula for the average speed of the speed boat.

$v_a=\frac{v_1{t}_1+v_2{t}_2}{t_1+t_2}$ (I)

Here, $v_a$ is the average speed, $v_1$ is the first velocity, $t_1$ is the time for which the speed is $v_1$, $v_2$ is the second velocity, $t_2$ is the time for which the speed is $v_2$

Conclusion:

Substitute $20.0\min$ for $t_1$, $10.0\min$ for $t_2$, $108\text{km/h}$ for $v_1$, $90.0\text{km/h}$ for $v_2$ in equation (I).

$\begin{array}{c}{v}_a=\frac{\left(108\text{km/h}\right)\left(20.0\min \right)+\left(90.0\text{km/h}\right)\left(10.0\min \right)}{20.0\min +10.0\min }\\ =102\text{km/h}\end{array}$

The average speed of the trip is $102\text{km/h}$.

(b)

To determine

The average velocity of the trip.

Answer to Problem 42P

The average velocity is $90.8\text{km/h}$ and directed $16.6°$ south of the west direction.

Explanation of Solution

The speed of the speed boat for first $20.0\min$ is $108\text{km/h}$ towards west. The speed of the boat for next $10.0\min$ is $90.0\text{km/h}$ towards $60.0°$ south of west.

Write the distance covered in west direction.

$d_1=v_1{t}_1$ (I)

Here, $d_1$ is the distance covered in west direction, $v_1$ is the first velocity, $t_1$ is the time for which the speed is $v_1$.

Write the distance covered in the direction towards $60.0°$ south of west.

$d_2=v_2{t}_2$ (II)

Here, $d_2$ is the distance covered in the direction towards $60.0°$ south of west, $v_2$ is the second velocity, $t_2$ is the time for which the speed is $v_2$.

The figure below shows the entire trip.

Write the formula for the x-component of the displacement.

$d_x=d_1+d_2\cos 240.0°$ (III)

Here, $d_x$ is the x-component of the displacement.

Write the formula for the y-component of the displacement.

$d_y=d_2\sin 240.0°$ (IV)

Here, $d_y$ is the y-component of the displacement.

Write the formula for the magnitude of the displacement.

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

Here, $d$ is the displacement.

Write the formula for the angle made by the displacement vector.

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

Here, $\theta$ is the angle made by the velocity vector.

Write the formula for the average velocity.

$\left|\overrightarrow{v}\right|=\frac{d}{t}$ (VII)

Here, $\overrightarrow{v}$ is the velocity vector, $t$ is the total time for the trip.

Conclusion:

Substitute $20.0\min$ for $t_1$, $108\text{km/h}$ for $v_1$ in equation (I).

$\begin{array}{c}{d}_1=\left(108\text{km/h}\right)\left(20.0\min \right)\\ =\left(108\text{km/h}\right)\left(20.0\min \right)\left(\frac{1\text{h}}{60\min }\right)\\ =36.0\text{km}\end{array}$

Substitute $10.0\min$ for $t_2$, $90.0\text{km/h}$ for $v_2$ in equation (II).

$\begin{array}{c}{d}_2=\left(90.0\text{km/h}\right)\left(10.0\min \right)\\ =\left(90.0\text{km/h}\right)\left(10.0\min \right)\left(\frac{1\text{h}}{60\min }\right)\\ =15.0\text{km}\end{array}$

Substitute $-36.0\text{km}$ for $d_1$, $15.0\text{km}$ for $d_2$ in equation (III).

$\begin{array}{c}{d}_x=-36.0\text{km}+\left(15.0\text{km}\right)\cos 240.0°\\ =-43.5\text{km}\end{array}$

Substitute $15.0\text{km}$ for $d_2$ in equation (IV).

$\begin{array}{c}{d}_y=\left(15.0\text{km}\right)\sin 240.0°\\ =-13.0\text{km}\end{array}$

Substitute $-43.5\text{km}$ for $d_x$, $-13.0\text{km}$ for $d_y$ in equation (V).

$\begin{array}{c}\left|d\right|=\sqrt{{\left(-43.5\text{km}\right)}^2+{\left(-13.0\text{km}\right)}^2}\\ =45.4\text{km}\end{array}$

Substitute $-43.5\text{km}$ for $d_x$, $-13.0\text{km}$ for $d_y$ in equation (VI).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-13.0\text{km}}{-43.5\text{km}}\right)\\ =16.6°\end{array}$

Substitute $45.4\text{km}$ for $d$, $30.0\min$ for $t$ in equation (VII).

$\begin{array}{c}\left|\overrightarrow{v}\right|=\frac{45.4\text{km}}{30.0\min }\\ =\frac{45.4\text{km}}{\left(30.0\min \right)\left(\frac{1\text{h}}{60\min }\right)}\\ =90.8\text{km/h}\end{array}$

The average velocity is $90.8\text{km/h}$ and directed $16.6°$ south of the west direction.