Question

Sally travels by car from one city to another. She drives for 30.0 min at 76.0 km/h, 39.0 min at 32.0 km/h, and 11.0 min at 33.0 km/h, and she spends 14.0 min eating lunch and buying gas.
(a) Determine the average speed for the trip?

Answer 1

Answer on Question #47214 – Physics - Mechanics | Kinematics | Dynamics

Sally travels by car from one city to another. She drives for 30.0 min at $76.0\mathrm{km/h}$, 39.0 min at $32.0\mathrm{km/h}$, and 11.0 min at $33.0\mathrm{km/h}$, and she spends 14.0 min eating lunch and buying gas. (a) Determine the average speed for the trip?

Solution:

Time $t_1 = 0.5$ hour at velocity $V_1 = 76 \frac{\text{km}}{\text{h}}$;

Time $t_2 = 0.65$ hour at velocity $V_2 = 32 \frac{\text{km}}{\text{h}}$;

Time $t_3 = 0.183$ hour at velocity $V_3 = 33 \frac{\text{km}}{\text{h}}$;

Time $t_4 = 0.23$ hour at velocity $V_4 = 0$;

The average speed is the total distance divided by the total travel time.

$$V_a = \frac{S}{t}$$

The total distance is

$$S = S_1 + S_2 + S_3 + S_4 = V_1 t_1 + V_2 t_2 + V_3 t_3 + V_4 t_4$$

The total time is

$$t = t_1 + t_2 + t_3 + t_4$$

(3) and (2) in (1):

$$V_a = \frac{V_1 t_1 + V_2 t_2 + V_3 t_3 + V_4 t_4}{t_1 + t_2 + t_3 + t_4} =$$ $$= \frac{0.5\, \text{h} \cdot 76\, \frac{\text{km}}{\text{h}} + 0.65\, \text{h} \cdot 32\, \frac{\text{km}}{\text{h}} + 0.183\, \text{h} \cdot 33\, \frac{\text{km}}{\text{h}} + 0.23\, \text{h} \cdot 0}{0.5\, \text{h} + 0.65\, \text{h} + 0.183\, \text{h} + 0.23\, \text{h}} = 41.5\, \frac{\text{km}}{\text{h}}$$

Answer: average speed for the trip is $41.5 \frac{\text{km}}{\text{h}}$

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