Question

Sally travels by car from one city to another. She drives for 25.0 min at 59.0 km/h, 50.0 min at 39.0 km/h, and 22.0 min at 40.0 km/h, and she spends 6.0 min eating lunch and buying gas.
(a) Determine the average speed for the trip.


km/h
(b) Determine the total distance traveled.


km

Answer 1

Answer on Question #46576, Physics, Mechanics | Kinematics | Dynamics

Problem.

Sally travels by car from one city to another. She drives for 25.0 min at 59.0 km/h, 50.0 min at 39.0 km/h, and 22.0 min at 40.0 km/h, and she spends 6.0 min eating lunch and buying gas.

(a) Determine the average speed for the trip.

km/h

(b) Determine the total distance traveled.

km

Solution:

$$25 \text{ min} = \frac{25}{60} \text{ h} = \frac{5}{12}, 50 \text{ min} = \frac{50}{60} \text{ h} = \frac{5}{6} \text{ h}, 22 \text{ min} = \frac{22}{60} \text{ h} = \frac{11}{30} \text{ h}, 6 \text{ min} = \frac{6}{60} \text{ h} = \frac{1}{10} \text{ h}.$$

The total time equals $t = 25 + 50 + 22 + 6 = 103$ min = $\frac{103}{60}$ h.

The distance equals speed multiplied by time.

Hence the total distance equals $S = 59 \cdot \frac{25}{60} + 39 \cdot \frac{50}{60} + 40 \cdot \frac{22}{60} + 0 \cdot \frac{6}{60} = 71.75$ km.

The average speed equals total distance divided by total time.

Hence the average speed equals $v = \frac{71.75}{\frac{103}{60}} \approx 41.7961 \, \text{km/h}$.

Answer: (a) $v = 41 \frac{82}{103} \approx 41.7961 \, \text{km/h}$. (b) $S = 71.75 \, \text{km}$.

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