Running and Walking: Distance, Rate, TimeDate: 06/22/99 at 22:14:50 From: Steve Thune Subject: Rate, distance and time Fred and Frank are two fitness fanatics on a run from A to B. Fred runs half the way and walks the other half. Frank runs half the time, and walks the other half. The two run at the same speed and walk at the same speed. Who finishes first? I think Frank, but I don't know how to prove it. Help! Date: 06/23/99 at 14:21:46 From: Doctor Rick Subject: Re: Rate, distance and time Hi, Steve. You're right. Here is an informal argument. Since Frank and Fred run faster than they walk, Frank will go farther in the time that he runs than in the equal time that he walks. Therefore he covers more distance at a run than Fred does, since Fred covers equal distances running and walking. While both are walking or both are running, they will cover equal distances. Therefore subtract this distance from the total distance. The remaining distance is covered by Fred at a walk and by Frank at a run, so Frank will finish first. Here is a graphical representation. Time runs up the graph. Fred runs (r) half the distance (covering more distance per unit of time), then walks (w), reaching B at the time marked Fred. Frank runs until the time marked "1/2 Frank," then walks until he reaches B at the time marked Frank. Frank's time is less. time ^ | Fred + w | w: | w : | w : | w : | w : Frank+ w w | Fred w w: | w w : | w w : | w w : | w w : | w w : | w w Frank: | w w : 1/2 +....................w.....r : Frank| w r : | w r : | r : | r : : | r : : | r : : | r : : | r : : +-----------------+-----------------+--> distance A 1/2 distance B You can formalize the argument by using algebra and the equation, distance = speed (rate) * time. Assign variables: speed while running = r speed while walking = w total distance = d Fred runs distance d/2; the time this takes is distance/speed = d/(2r). He walks distance d/2; the time this takes is d/(2w). Thus Fred's total time is t_Fred = d/(2r) + d/(2w) = d(r + w)/(2rw) Frank runs for time t_Frank/2, covering distance = speed * time = r*t_Frank/2. He walks for time t_Frank/2, covering distance w*t_Frank/2. The total distance is d = r*t_Frank/2 + w*t_Frank/2 = (r + w)t_Frank/2 Solving for t_Frank, t_Frank = 2d/(r + w) Which is greater? Subtract t_Frank from t_Fred and see whether it is positive. t_Fred - t_Frank = d(r+w)/(2rw) - 2d/(r+w) = d((r+w)^2 - 4rw)/(2rw(r+w)) = d(r^2 - 2rw + w^2)/(2rw(r+w)) = d(r-w)^2/(2rw(r+w)) Since d, (r-w)^2, r, w, and (r+w) are all positive, this difference is positive. Therefore t_Fred > t_Frank Frank takes less time than Fred, and finishes first. So now you have proof. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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