A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min.
(a) What was her average speed?
(b) If the straight-line distance from her home to the university is 10.3 km in a direction 25° S of E, what was her average velocity?
(c) If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip?
Solution:
Part A
The average speed is
\begin{align*}
\text{speed} & = \frac{\text{distance}}{\text{time}}\\
&= \frac{12\:\text{km}}{18\:\text{mins}}\times \frac{60\:\text{mins}}{1\:\text{hr}} \\
& =40\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B
The average velocity is
\begin{align*}
\overline{v} & =\frac{\Delta \text{x}\:}{\Delta \text{t}} \\
& =\frac{10.3\:\text{km}}{18.0\:\min \:}\times \frac{60\:\text{mins}}{1\:\text{hr}} \\
&=34.33\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}The direction of the velocity is 25° S of E \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right).
Part C
The average speed is
\begin{align*}
\text{speed} & = \frac{\text{distance}}{\text{time}}\\
& =\frac{12.0\:\text{km}\times 2}{7.5\:\text{hr}} \\
& =3.2\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}And the average velocity is
\begin{align*}
\overline{v}=0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}The average velocity is zero since the total displacement is zero.
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