Question

Sally travels by car from one city to another. She drives for 21.0 min at 76.0 km/h, 53.0 min at 40.0 km/h, and 49.0 min at 49.0 km/h, and she spends 6.0 min eating lunch and buying gas. Determine the average speed for the trip.

Answer 1

Answer on Question #45619, Physics, Mechanics | Kinematics | Dynamics

Sally travels by car from one city to another. She drives for 21.0 min at $76.0\mathrm{km/h}$, 53.0 min at $40.0\mathrm{km/h}$, and 49.0 min at $49.0\mathrm{km/h}$, and she spends 6.0 min eating lunch and buying gas. Determine the average speed for the trip.

Solution:

The average speed during the course of a motion is often computed using the following formula:

$$\text{erage Speed} = \frac{\text{Distance Traveled}}{\text{Time of Travel}}$$ $$v_{av} = \frac{d_1 + d_2 + d_3 + d_4}{t_1 + t_2 + t_3 + t_4}$$ $$v_1 = 76.0\ \mathrm{km/h},$$ $$v_2 = 40.0\ \mathrm{km/h},$$ $$v_3 = 49.0\ \mathrm{km/h},$$ $$v_4 = 0,$$ $$t_1 = 21\ \mathrm{min} = \frac{21}{60} = 0.35\ \mathrm{hour},$$ $$t_2 = 53\ \mathrm{min} = \frac{53}{60} = 0.883\ \mathrm{hour},$$ $$t_3 = 49\ \mathrm{min} = \frac{49}{60} = 0.817\ \mathrm{hour},$$ $$t_4 = 6\ \mathrm{min} = \frac{6}{60} = 0.1\ \mathrm{hour},$$ $$d_1 = v_1 t_1 = 76 \cdot 0.35 = 26.6\ \mathrm{km}$$ $$d_2 = v_2 t_2 = 40 \cdot 0.883 = 35.32\ \mathrm{km}$$ $$d_3 = v_3 t_3 = 49 \cdot 0.817 = 40.033\ \mathrm{km}$$ $$d_4 = v_4 t_4 = 0\ \mathrm{km}$$

The average speed

$$v_{av} = \frac{26.6 + 35.32 + 40.033 + 0}{0.35 + 0.883 + 0.817 + 0.1} = 47.42\ \mathrm{km/h}$$

Answer: $v_{av} = 47.4\ \mathrm{km/h}$

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