Asked by JP
Stan takes a trip, driving with a constant velocity of 81.5 km/h to the north except for a 22 min rest stop. If Stan's avg velocity is 66.80 km/h to the north, how long does the trip take?
Answers
Answered by
Steve
Let the driving time be t hours.
The distance covered is 81.5t km.
However, if we factor in the rest stop of 22/60 hours, the distance is 66.8(t+22/60).
Equating the two distances, we get
66.8(t+22/60) = 81.5t
66.8t + 25 = 81.5t
25 = 14.7t
t = 1.7 hours driving, so
the trip took 1hr 42min + 22min = 2hr 4min
The distance covered is 81.5t km.
However, if we factor in the rest stop of 22/60 hours, the distance is 66.8(t+22/60).
Equating the two distances, we get
66.8(t+22/60) = 81.5t
66.8t + 25 = 81.5t
25 = 14.7t
t = 1.7 hours driving, so
the trip took 1hr 42min + 22min = 2hr 4min
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