Race between a Speedy Tortoise and a Resting Hare

AI Thread Summary
The discussion revolves around solving a race problem involving a speedy tortoise and a hare that stops to rest. The tortoise runs at 10 cm/s, while the hare runs at 200 cm/s but takes a 2-minute break. To find the total race time and length, participants suggest setting up equations based on their speeds and distances, with the tortoise finishing 20 cm ahead. The key equations involve the tortoise's time to complete the race and the hare's adjusted running time after resting. The conversation emphasizes solving these equations to determine the race's duration and distance.
scottzilla
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Homework Statement



Speedy tortoise can run 10cm/s, and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 2 minutes. The tortoise wins the race by a shell (20cm). A)How long does the race take? B)What is the length of the race.

Homework Equations





The Attempt at a Solution


For some reason I am lost on how to set up this problem. I keep wanting to use motion in 1 dimension with constant acceleration but I cannot come up with the answers given in the book. A step by step explanationj on how to solve this type of question would be awesome.

Scottzilla
 
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This question is probably a little simpler than you're trying to make it.

Okay, let's start by writing down what we know. Suppose that the length of the track is L and the time taken for the tortoise to complete that race, T, such that

v = \frac{dx}{dt}\Rightarrow 10 = \frac{L}{T}

Next we have the hare's information. We know that the hare stops of 2 minutes or 120 seconds, so the total time the hare is running is T-120. We also know that when the tortoise is at the finish line (x=L), the hare is 20cm behind him, i.e. x=L-20. Finally, we know that the hare can run at 20x10 cm/s. Hence,

v = \frac{dx}{dt}\Rightarrow 200 = \frac{L-20}{T-120}

Do you follow?
 
Ok I think I follow you. When I was trying to come up with my equations I was taking the hare's distance to be equal to the tortoise's distance minus 20cm. SO should I take those equations and solve one for (T)ime and then put that equation into the other one?
 
scottzilla said:
When I was trying to come up with my equations I was taking the hare's distance to be equal to the tortoise's distance minus 20cm.
That's exactly what I've done with my set of equations, L-20
scottzilla said:
SO should I take those equations and solve one for (T)ime and then put that equation into the other one?
Sounds good to me :approve:
 
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