Question

Sally travels by car from one city to another. She drives for 28.0 min at 60.0 km/h, 38.0 min at 31.0 km/h, and 22.0 min at 35.0 km/h, and she spends 14.0 min eating lunch and buying gas.
(a) Determine the average speed for the trip.

(b) Determine the total distance traveled.

Answer 1

Question 35246

Let $t_i$ denote time of moving through interval $L_i$ with velocity $v_i$ . Using this notation, $t_1 = 0.5h$ ; $v_1 = 80\frac{km}{h}$ ; $t_2 = \frac{12}{60}h$ ; $v_2 = 105\frac{km}{h}$ ; $t_3 = \frac{45}{60}h$ ; $v_3 = 40\frac{km}{h}$ . Also, time spent for buying gas is $t' = \frac{21}{60}h$ .

a)

The average speed is the total distance divided by time it took to cover this distance, $v = \frac{L}{t}$ . In this case, time is sum of three times moving on three intervals plus time needed to buy gas:

$t = t_{1} + t_{2} + t_{3} + t^{\prime} = 0.5 + \frac{12}{60} +\frac{45}{60} +\frac{21}{60} = \frac{9}{5} h$ . Total distance is $L = v_{1}t_{1} + v_{2}t_{2} + v_{3}t_{3} = 91km$ . Hence, average velocity is $v = \frac{91km}{\frac{9}{5}h} = \frac{455}{9}\frac{km}{h}\approx 50.56\frac{km}{h}$

b) The total distance traveled is already calculated in a):

$$L = v _ {1} t _ {1} + v _ {2} t _ {2} + v _ {3} t _ {3} = 0. 5 \cdot 8 0 + \frac {1 2}{6 0} \cdot 1 0 5 + \frac {4 5}{6 0} \cdot 4 0 = 9 1 k m.$$