Piece of Cake Stumper Answer 12 November 1999 http://www.rain.org/~mkummel/05nov99a.html Copyright (c) 1999 by Marc Kummel aka Treebeard. Contact mkummel@rain.org, http://www.treebeard.org/ ---------------------------------------------------------------------------- The Dunn Middle School Piece of Cake bike ride this weekend is a slow ride up the big hill along Santa Rosa Creek from Cambria, followed by a fast downhill to Morro Bay. I don't know the actual distances up and down, but they're close, so we can assume they're equal for this stumper. It's steep, so you pedal (and push) up the hill at an average speed of just 5 miles per hour. Then you zoom down to the end as fast as you can. How fast do you have to ride down so that your average speed for the whole trip is 10 miles per hour? Careful, this is a stumper! ---------------------------------------------------------------------------- I ride up the hill at just 5 miles per hour. But even if I ride the same distance down the hill at the speed of light, it's not fast enough for me to average 10 mph for the whole ride! The distance doesn't matter for this problem, so pick an easy one, say 20 miles each way. At 5 mph, it takes 4 hours to ride the 20 miles up. But to average 10 mph, I must ride the whole 40 miles in (40 miles / 10 mph) = 4 hours. That's impossible since I've already used up all my time. This stumper shows how even familiar concepts can contain unexpected loopholes! Notes: Without exception, the DMS kids guessed I have to ride down the hill at 15 miles per hour. That answer seems so easy since (5 + 15) / 2 = 10. But it's wrong: d 20 miles Time to ride up = -- = -------- = 4 hours v1 5 mph d 20 miles Time to ride down = -- = -------- = 1.3 hours v2 15 mph d 40 miles Average speed = - = --------- = 7.5 miles per hour, not 10! t 5.3 hours This problem is tricky because we're really working with an average of averages. Graybear sent this answer: The speed o' light! What can I say? If you averaged 6 mph up and wanted to average 10 mph for the whole trip, I would have had to think about it a minute. (30 mph) How about creating a graph of speed up the hill vs. speed down to help explain the math? Use the familiar formulas d=vt, v=d/t, and t=d/v for this problem. Say my speed up the Piece of Cake hill is 6 miles per hour, and the distance is 20 miles up and 20 miles down (though it really doesn't matter). I want to average 10 miles per hour for the whole trip, so I must ride the whole way in (2 * 20 miles / 10 mph) = 4 hours. I figure that I spend (20 miles / 6 mph) = 3.3 hours pedaling up. That leaves me (4 - 3.3) = 0.7 hours to ride the 20 miles downhill. So my downhill speed is d / t = (20 miles / 0.7 hours) = 30 miles per hour. We can generalize this with some simple algebra. d = distance up (and down) the hill, so the total distance is 2d t = time, with separate t1 (up), t2 (down), and t (total) v = my average speed for the whole ride v1 = my average speed up the hill v2 = my average speed down the hill My time to go up the hill is: d t1 = -- v1 To achieve my desired average speed, I must do the whole ride in: 2d t = -- v So the time remaining to ride the downhill is: 2d d t2 = t - t1 = -- - -- v v1 And so my downhill speed must be: d d v2 = ---- = --------- t2 2d d -- - -- v v1 We can simplify this by applying the distributive law in the denominator to pull out the common factor d: d = ------------- 2 1 d ( - - -- ) v v1 The distance d now cancels out. It doesn't matter what the distance is for this problem! 1 = -------- 2 1 - - -- v v1 1 = ------------- 2v1 - v ------- v v1 So we finally arrive at a general solution: v v1 v2 = ------- 2v1 - v Look at the term under the division line. If my average velocity for the whole trip (v) is greater than or equal to my average uphill speed (v1), then my downhill speed (v2) is negative or undefined. You can't average twice your average speed for the first half of any distance. Q.E.D. ---- Last modified 13 November 1999.