Asked by John
You jog at 6.2 mi/h for 5.0 mi, then you jump into a car and drive for another 5.0 mi. With what average speed must you drive if your average speed for the entire 10.0 miles is to be 10.6 mi/h?
I don't get why the answer isn't 15.
Answers
Answered by
tchrwill
The average speed for the entire trip is the total distance divided by the total time.
Therefore,
t1 = 5/6.2 = .806 hr.
t2 = 5/V
Total distance traveled is 5 + 5 = 10 miles.
Total time is t1 + t2 = .806 + 5/V
Average driving speed is therefore
Vavg = 10/(.806 + 5/V) = 10.6
Solving,V = 36.511mph.
(5 + 5)/(.806 + 5/36.511) = 10.6mph.
Therefore,
t1 = 5/6.2 = .806 hr.
t2 = 5/V
Total distance traveled is 5 + 5 = 10 miles.
Total time is t1 + t2 = .806 + 5/V
Average driving speed is therefore
Vavg = 10/(.806 + 5/V) = 10.6
Solving,V = 36.511mph.
(5 + 5)/(.806 + 5/36.511) = 10.6mph.
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