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# Treebeard's Stumper Answer22 January 1999

Up This Hill and Down

Here's a stumper to ponder as you pedal up the big hill on our annual Dunn Middle School Piece of Cake bike ride this weekend. Betty and Carl start riding at the same time and don't stop. Betty paces herself and does the whole ride at a constant speed. Carl is slow going uphill, so he covers the first half of the distance at a speed 5 miles per hour slower than Betty. He makes up for it coming down, and rides the second half at a speed 5 miles per hour faster than Betty. Who finishes the ride first? Or do they arrive at the same time? It's a piece of cake!

Carl starts slow and finishes fast, while Betty keeps a pace. Since no particular speed or distance is given, pick easy ones. Suppose the ride is 30 miles and Betty bikes at 10 mph. She finishes in 3 hours. Carl does the first 15 miles at 5 mph in 3 hours. He rides the second 15 miles in just an hour, but Betty is already there. The result is the same for any speed and distance. Aesop says that slow and steady wins the race. On our ride last weekend, the Saturn pro racing team passed us on the road. Fast and steady really wins races!

Notes:

Rain was threatening, so we canceled our traditional DMS Piece of Cake bike ride from Cambria to Morro Bay. Instead we rode along Foxen Canyon Road from school to the historic San Ramon Chapel in the small town of Sisquoc. I took pictures of our ride. At the Chapel, we visited the grave of the English pioneer Benjamin Foxen who became part of California history by helping Gen. John C. Fremont take Santa Barbara from Mexico for the new California Republic.

According to the marker on his grave, Foxen was born in 1796 and died in 1874. "Don Julian" Foxen came to California in 1818. Here's another stumper: How many nations did Foxen live under while in California? (hint) Is there any other place in the world where you could visit more countries by just staying in the same place?

Here's the original Aesop story of The Hare and the Tortoise (from Project Gutenberg):

A HARE one day ridiculed the short feet and slow pace of the Tortoise, who replied, laughing: "Though you be swift as the wind, I will beat you in a race." The Hare, believing her assertion to be simply impossible, assented to the proposal; and they agreed that the Fox should choose the course and fix the goal. On the day appointed for the race the two started together. The Tortoise never for a moment stopped, but went on with a slow but steady pace straight to the end of the course. The Hare, lying down by the wayside, fell fast asleep. At last waking up, and moving as fast as he could, he saw the Tortoise had reached the goal, and was comfortably dozing after her fatigue.

Slow but steady wins the race.

It's easy to prove the general case of this week's stumper with a bit of algebra. To remember the formula for distance, velocity, and time, use Treebeard's Rule: Pick an easy case that you can do without thinking, see what you did, and generalize. If I go 10 miles (d) at 5 miles per hour (v), then it takes me 2 hours (t). 2 is 10 / 5, so t = d / v.

Suppose the distance up (and down) is d, so the whole ride is distance 2d. Betty's constant speed is v. Carl rides s units slower up and faster down than Betty.

It takes Betty time d/v to get up and down, for a total of 2d/v for the whole ride. Multiplying top and bottom by v gives:

```              2dv
t_betty = ---
v^2
```

Carl's speed on the way up is v-s, and v+s on the way down. So it takes Carl time d/(v-s) to get up, and d/(v+s) to get down, for a total of:

```               d     d    d(v+s) + d(v-s)   dv + ds + dv - ds      2dv
t_carl  = --- + --- = --------------- = ----------------- =  --------
v-s   v+s      (v+s)(v-s)           v^2-s^2         v^2-s^2
```

If the bottom (denominator) of a fraction is made greater, then the value of the fraction must become smaller. 1/3 is less than 1/2 since 3 is greater than 2. Therefore (for v>s and s>0),

```        2dv     2dv
--- < -------
v^2   v^2-s^2
```
So t_betty < t_carl.
Betty wins the race everytime.

It's harder to explain in words. I figure that Carl is slower going up, and he's slower for longer. The faster (and quicker) downhill doesn't compensate. It's different if Carl rides the two halves of the time of his ride slower and faster.