200 ft race (with turnaround) where two people have different jump lengths but jump the same distance in equal time - who wins and by how much?

Question

This problem is from Problem Solving Strategies - Crossing the River with Dogs and Other Mathematical Adventures by Ken Johnson and Ted Herr.

I decided to draw a picture from the segment of $90$ ft to $100$ ft of the race (the wall where they turn around). I chose to start at $90$ ft because both two and three go into 90 evenly, so neither Roo or Tigger are in the middle of a jump at that time.

Thus, at the $96$ foot mark, both Roo and Tigger are at the same spot. But, at the $98$ foot mark, Roo is about to jump to the $100$ ft mark and Tigger is mid-jump en route to the $99$ ft mark. So, when Roo jumps, he is at the $100$ ft mark and ready to turn around, while Tigger is on the $102$ ft mark and then waiting to turn back around.

I am having trouble trying to figure out how much Roo wins by then exactly. It's hard to tell from my picture, but I believe it would be $2$ feet. Is that correct, or does Roo possibly win by more? Any help would be appreciated.

Answer 1

Roo travels $200$ feet to reach the finish line.

Since Tigger has to travel in $3$ foot jumps, Tigger needs to travel $102$ feet until (s)he can turn around. Then Tigger needs to travel another $102$ feet to reach the start/finish line, a total of $204$ feet. The two jumpers travel at the same rate. So Roo is $4$ feet ahead of Tigger at the end, sort of.

This is in fact not necessarily the case, because it relies on a specific model of jumping. Since we were not given a detailed description of the jump processes of either creature, we are forced to assume something. We implicitly assumed that even although the travelling is divided into jumps, the actual speeds are constant.

Suppose instead that a jump consists of standing in one place flexing the leg muscles for a while, and perhaps having a drink, while the movement part of a jump is quick. Then the answer need not be $4$ feet.

Answer 2

You can also solve it by counting jumps. Roo reaches the halfway point in exactly $50$ jumps, so he requires $100$ jumps to finish the race. Tigger reaches the $99$ foot mark in $33$ jumps, then takes $2$ more jumps to cross the $100$ foot mark and return to the $99$ foot mark, and finally needs $33$ jumps to return from there to the starting line, for a total of $68$ jumps. At $3$ jumps per second (or whatever the basic unit of time is), Roo finishes in $\frac{100}3$ seconds. In $\frac{100}3$ seconds Tigger makes $2\cdot\frac{100}3=\frac{200}3$ jumps, which is $\frac43$ of a jump, or $4$ feet, short of the $68$ jumps that he requires. Assuming that his speed is constant, he’s still $4$ feet shy of the finish line when Roo crosses it.