   SEARCH HOME Math Central Quandaries & Queries  Question from jenna, a student: A tortoise can run with a speed of 0.14 m/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 1.0 minutes. The tortoise wins by a shell (25 cm). (a) How long does the race take? (b) What is the length of the race? Hi Jenna.

distance (d) = speed (s) times time (t).

In this case, the distance is 25cm further for the Tortoise than for the Hare.

So we can use subscripts to identify the relationships:

dH = sH tH

dT = sT tT

We also know how these things are related to each other.

First, the Tortoise beat the Hare by 25 cm. So

dH = dT - 0.25

Also, the speed of the Hare was 20 times the speed of the Tortoise:

sH = 20 sT

Finally, we also know that the Hare took 60 seconds off for a rest:

tH = tT - 60

If you look at this set of equations, you will see that we can substitute these last three equations into the first one! So

dH = sH tH

becomes

dT - 0.25 = (20 sT) (tT - 60)

Lastly, we know that sT = 0.14 m/s, so we can substitute that into both this last equation and the second one. Since everything left is subscripted with a T, I'm going to leave it off. We are left with two equations with two unknowns:

d - 0.25 = (20) (0.14) (t - 60)
d = (0.14) t

You can use either the substitution method or the elimination method to solve these simultaneous equations (look those up in our Quick Search if you don't know them). I suggest eliminating d and then you can find t, the time the Tortoise took to win rthe race. Then you can use that to quickly find d, the length of the race.

Hope this helps you,
Stephen La Rocque     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.