Q. 83 - Precalculus Enhanced with Graphing Utilities

Question

The Tortoise and the Hare: In a 21-meter race between a tortoise and a hare, the tortoise leaves 9 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 3 minutes before the tortoise. What are the average speeds of the tortoise and the hare?

Step-by-Step Solution

Answer

The speed of the tortoise is 7 meters per second and the speed of the hare is 7.5 meters per second.

Step 1. Find the equation for the average speed of the tortoise and hare.

Let the average speed of the tortoise be x and that of the hare be y. 

Given that the speed of the hare is 0.5 meter per hour faster than that of the tortoise.

which gives

$y=x+0.5$

Therefore, the average speed of the hare is x+0.5 and that of the tortoise is x.

Step 2. Use the relation of the speed, distance and time.

The total distance covered by both the hare and the tortoise is 21 meters.

So, the time taken by the tortoise to cover the distance is $\frac{21}{x}$ and the time taken by the hare is $\frac{21}{y}$

Substitute $x+0.5$ for y in $\frac{21}{y}$.

$\frac{21}{y}=\frac{21}{x+0.5}$

Step 3. Use the data for the time given.

It is given that the tortoise leaves the starting point 9 minutes before the hare and reaches the finishing point 3 minutes after the hare.

So, the tortoise is slower by 12 minutes than the hare.

Since the tortoise takes 12 minutes more than the hare to reach the finishing point, the total time taken will be given by $\frac{21}{x+0.5}+12$ minutes and

$\frac{12}{60}=\frac{1}{5}$ hours which gives

$\frac{21}{x+0.5}+\frac{1}{5}=\frac{21}{x}$

Step 4. Solve the equation 21 x + 0 . 5 + 1 5 = 21 x .

We get

$$\begin{gathered} \frac{21}{x+0.5}+\frac{1}{5}-\frac{21}{x+0.5}=\frac{21}{x}-\frac{21}{x+0.5} \\ \frac{1}{5}=\frac{21}{x}-\frac{21}{x+0.5} \\ \frac{1}{5}=21(\frac{1}{x}-\frac{1}{x+0.5}) \\ \frac{1}{5}=21\left(\frac{x+0.5-x}{x(x+0.5)}\right) \\ \frac{1}{5}=21\left(\frac{0.5}{x(x+0.5)}\right) \\ \frac{1}{5}=\frac{10.5}{x(x+0.5)} \\ 1=\frac{10.5}{x(x+0.5)}\times 5 \\ 1=\frac{52.5}{x(x+0.5)} \\ x(x+0.5)=52.5 \\ {x}^2+0.5x-52.5=0 \end{gathered}$$

Using quadratic equation formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$,

$$\begin{gathered} x=\frac{-0.5\pm \sqrt{{(0.5)}^2-4\left(1\right)(-52.5)}}{2\left(1\right)} \\ =\frac{-0.5\pm \sqrt{0.25+210}}{2} \\ =\frac{-0.5\pm \sqrt{210.25}}{2} \\ =\frac{-0.5\pm 14.5}{2} \end{gathered}$$

which means

$$\begin{gathered} x=\frac{-0.5-14.5}{2} or x=\frac{-0.5+14.5}{2} \\ x=-7.5 or x=7 \end{gathered}$$

Step 5. Find the required speed.

We discard $-7.5$ as speed cannot be negative.

So, $x=7$ and

$$\begin{gathered} y=x+0.5 \\ =7+0.5 \\ =7.5 \end{gathered}$$

The speed of the tortoise is 7 meters per second and the speed of the hare is 7.5 meters per second.