Race between a Speedy Tortoise and a Resting Hare

[scottzilla]

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

 

[Hootenanny]

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?

 

[scottzilla]

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?

 

[Hootenanny]

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

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