The Tortoise and the Hare . . . Revisited(2)

This month's puzzle is a new version of an old story.

There’s an old story about a race between a tortoise and a hare. The hare knew he was much faster than the tortoise. He was so confident of winning that he lay down by the side of the road to take a nap. Meanwhile the tortoise just kept going, slowly and steadily toward the finish line.

The Tortoise and the Hare

The hare overslept. When he finally woke up and sped toward the finish, it was too late to catch up. The moral of the story is “slow and steady wins the race.”

Our hare has heard that story many times, so he came away with a different lesson: “Always set your alarm clock.”

On the morning of a 10 km (6.2 mi) race, the hare and the tortoise were at the starting line. As the starting whistle blew, the tortoise set off down the road at top speed. For him that was 1 km/h (.62 mph). The hare set his alarm clock and went to sleep.

The hare was able to run at 5 km/h (3.1 mph). How much of a nap could he take and still win?

Can you come up with a general solution to this problem that will tell the hare how long a nap he can take for any combination of hare speed, tortoise speed, and race length?

Course:

Math [1]
Algebra [2]

Result/Solution(s)

Solution: The Tortoise and the Hare . . . Revisited Math Puzzle

The tortoise can run the 10-km course in 10 hours. The hare can do it in 2 hours. So to tie, he has to get going when the tortoise has 2 hours left. This is when the tortoise has run 8 km (5 mi) and has 2 km (1.2 mi) to go, so the hare can sleep for 8 hours, minus a few seconds to guarantee the win.

The Tortoise and the Hare

Let’s come up with a general solution. Let’s say

T = tortoise speed
H = hare speed
R = length of the race

Let’s assume that the hare is faster.

The time for the tortoise to run the race is

R / T

The time for the hare to run the race is

R / H

The hare has to get going when time elapsed in the race is

(R / T) – (R / H)

In our example we would have

(10 / 1) – (10 / 5) = 10 – 2 = 8

For a tie, the hare needs to start after the tortoise has been running for 8 hours.

Let’s say the race is 15 km (9.3 mi) and the two animals run at the same speeds as in the example above. We have

(15 / 1) – (15 / 5) = 15 – 3 = 12

The hare can nap for 12 hours.

If the tortoise runs at 2 km/h and the hare still runs at 5 km/h, a 10-km race looks like this:

(10 / 2) – (10 / 5) = 5 – 2 = 3

With this speedier tortoise, the hare gets only a 3-hour nap.

Our formula should work for any combination of tortoise speed, hare speed, and race length. The answers we calculated are for a tie between the two competitors. The hare would have to set his alarm clock slightly ahead of the specified time to win.