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In a 21 meter race between a tortoise and a hare, the tortoise leaves 8 minutes before the hare. The hare by running at an average speed of 0.5 meter per hour faster than the tortoise, crosses the finish line 4 minutes before the tortoise. What are the average speeds of the tortoise and the hare?

In a 21 meter race between a tortoise and a hare, the tortoise leaves 8 minutes before-example-1

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Explanation:

Let's call the average speed of the tortoise "t" (in meters per hour) and the average speed of the hare "h" (in meters per hour).

From the problem, we know that:

The tortoise leaves 8 minutes before the hare, so they have a 4-minute head start.

The hare crosses the finish line 4 minutes before the tortoise, so they have a 4-minute lead.

Therefore, the total time it takes for the hare to finish the race is 8 minutes less than the time it takes for the tortoise to finish the race. Let's call this time difference "dt".

The distance the hare runs is 21 meters, and the distance the tortoise runs is also 21 meters, so we have:

h * dt = 21 - t * (dt + 4 minutes)

To solve for the average speeds "t" and "h", we need to convert everything to units of hours. Let's convert 4 minutes to hours:

4 minutes = 4/60 hours = 1/15 hours

So, we can now rewrite the equation in terms of hours:

h * dt = 21 - t * (dt + 1/15 hours)

Rearranging and solving for t, we find:

t = (21 + h * dt) / (dt + 1/15)

Now, the hare runs 0.5 meters per hour faster than the tortoise, so:

h = t + 0.5

Substituting h = t + 0.5 into the equation for t, we get:

t = (21 + (t + 0.5) * dt) / (dt + 1/15)

Solving for t, we find:

t = 40/3 meters per hour

Finally, we can find the average speed of the hare by using h = t + 0.5:

h = 40/3 + 0.5 = 40/3 + 30/60 = 40/3 + 30/3 / 60 = 70/3 meters per hour

So the average speed of the tortoise is 40/3 meters per hour, and the average speed of the hare is 70/3 meters per hour

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