Question

Sally travels by car from one city to another. She drives for 28.0 min at 79.0 km/h, 49.0 min at 31.0 km/h, and 49.0 min at 70.0 km/h, and she spends 10.0 min eating lunch and buying gas.
Determine the average speed for the trip.

Answer 1

Question 34506

The average speed is given by $\nu_{\text{average}} = \frac{S}{t}$ , where $S$ is the distance covered, and $t$ is the time needed to cover this distance.

Sally travels three different distances at different speeds. Each of the distance is $l = v \cdot t$ , where $v$ is the speed on particular part of the road and $t$ is the time she moved with that speed. Hence, total distance is $S = v_1 t_1 + v_2 t_2 + v_3 t_3 = \frac{28}{60} h \cdot 79 \frac{km}{h} + \frac{49}{60} h \cdot 31 \frac{km}{h} + \frac{49}{60} h \cdot 70 \frac{km}{h} = 119.35 \, km$ . Total time of the trip is $t_1 + t_2 + t_3 + T = \frac{28 + 49 + 49}{60} + \frac{10}{60} \approx 2.267 \, h$ (Three time intervals needed to cover three corresponding distances plus 10 minutes for eating lunch and buying gas).

Thus, average speed is $\nu_{\text{average}} = \frac{119.35 \, \text{km}}{2.267 \, h} \approx 52.65 \, \frac{\text{km}}{\text{h}}$ .