Question #14911

A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 50.0 min at 95.0 km/h, 7.0 min at 80 km/h, and 50.0 min at 55.0 km/h, and spends 30.0 min eating lunch and buying gas. '
a. determine the average speed for the trip
b.Determine the distance between the initial and final cities along this route.

Expert's answer

a. determine the average speed for the trip

The total distance traveled is

L = 95/60[km/m]*50[m] + 80/60[km/m]*7[m] + 55/60[km/m]*50[m] = 134+1/3 [km].

The total time elapsed is

T = 50[m] + 7[m] + +50[m] + 30[m] = 137 [m].

So, the average speed for the trip was

V = L/T = (134+1/3)[km] / (137)[m] ≈ 0.9805 [km/m] = 58.83 [km/h].

&

b.Determine the distance between the initial and final cities along this route.

The distance between the initial and final cities is L = 134+1/3 [km].

The total distance traveled is

L = 95/60[km/m]*50[m] + 80/60[km/m]*7[m] + 55/60[km/m]*50[m] = 134+1/3 [km].

The total time elapsed is

T = 50[m] + 7[m] + +50[m] + 30[m] = 137 [m].

So, the average speed for the trip was

V = L/T = (134+1/3)[km] / (137)[m] ≈ 0.9805 [km/m] = 58.83 [km/h].

&

b.Determine the distance between the initial and final cities along this route.

The distance between the initial and final cities is L = 134+1/3 [km].

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