recall your d = rt, distance = rate * time.
w = speed of the wind.
14 = speed of the bicycle without wind.
now, against the wind, the bicycle is not really going 14 mph fast, is really going "14 - w", since the wind is eroding speed from it, likewise, when the bicycle is going with the wind is not going 14 mph fast either, is really going "14 + w" due to the wind adding speed, let's say it took "t" hours against and also "t" hours with it.

