Chapter 3, Problem 30P

(a)

To determine

The average speed of the trip.

Answer to Problem 30P

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

Explanation of Solution

Write the expression for the average speed of the 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  $\text{20}\text{.0min}$ for $t_1$, $108\text{km/h}$ for $v_1$ , $90.0\text{km/h}$ for $v_2$ , and $\text{10}\text{.0min}$ for $t_2$ in expression (I),

  $\begin{array}{c}{v}_a=\frac{\left(108\text{km/h}\right)\left(\text{20}\text{.0min}\right)+\left(90.0\text{km/h}\right)\left(\text{10}\text{.0min}\right)}{\text{20}\text{.0min}+\text{10}\text{.0min}}\\ =102\text{km/h}\end{array}$                                      

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

(b)

To determine

The average velocity of the trip.

Answer to Problem 30P

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

Explanation of Solution

Write the expression for the distance covered in west direction,

  $d_1=v_1{t}_1$        (I)

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

Write the expression for the distance covered in the direction towards $60.0°$ south west,

  $d_2=v_2{t}_2$        (II)

Here, $d_2$ is the distance, $v_2$ is the first velocity, $t_2$ is the time for which the speed is $v_2$.

The figures 1 the total 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 displacement along $x$ component

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

  $d_y=d_2\sin 240.0°$        (IV)

Here, $d_y$ is the displacement along $y$ component

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.

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

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

Write the expression for the average velocity

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

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

Conclusion:

Substitute  $\text{20}\text{.0min}$ for $t_1$, $108\text{km/h}$ for $v_1$ in expression (I),

  $\begin{array}{c}{d}_1=\left(108\text{km/h}\right)\left(\text{20}\text{.0min}\right)\\ =\left(108\text{km/h}\right)\left(\text{20}\text{.0min}\right)\left(\frac{\text{1h}}{\text{60min}}\right)\\ =36.0\text{km}\end{array}$                                      

Substitute  $90.0\text{km/h}$ for $v_2$ , and $\text{10}\text{.0min}$ for $t_2$ in expression (II)

  $\begin{array}{c}{d}_2=\left(90.0\text{km/h}\right)\left(\text{10}\text{.0min}\right)\\ =\left(90.0\text{km/h}\right)\left(\text{10}\text{.0min}\right)\left(\frac{\text{1h}}{\text{60min}}\right)\\ =15.0\text{km}\end{array}$

Substitute  $-36.0\text{km}$ for $d_1$, $15.0\text{km}$ for $d_2$ in expression (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 expression (IV)

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

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

  $\begin{array}{c}\left|d\right|=\sqrt{{\left(-13.0\text{km}\right)}^2+{\left(-43.5\text{km}\right)}^2}\\ =45.5\text{km}\end{array}$

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

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

Substitute  $45.5\text{km}$ for $d$ , and $\text{30}\text{.0min}$ for $t$ in expression (VII)

  $\begin{array}{c}\left|\overrightarrow{v}\right|=\frac{45.5\text{km}}{\text{30}\text{.0min}\left(\frac{\text{1h}}{\text{60min}}\right)}\\ =90.\text{8km/h}\end{array}$

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