Formula for Necessary Return Speed

If Alice can make a trip of m miles in t hours, at what rate must she make the return trip, given that it is possible to do so, so that the average speed for the whole journey will be k miles per hour?

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We’ve got a lot of variables already, but we go ahead and let x represent the rate for the return trip.

Then the usual chart:

  d = r t
go m m/t t
come m x m/x
avg. 2m k t + m/x

Hmm: What if 2m = kt? Well, suppose the distance m = 10 miles and the time t = 2 hours, so the “going” rate is 10/2 or 5 miles per hour. Then 2(10) = k · 2 → the desired average is 10 mi/hr. But going 2m = 20 miles at an average speed of 10 mi/hr will take 2 hours, and going already took 2 hours, so there is no time to get back. Thus, we can never ask that the average speed be double the “going” rate.

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