Q. 84 - Precalculus Enhanced with Graphing Utilities

Question

Running a Race : In a 1-mile race, the winner crosses the finish line 10 feet ahead of the second-place runner and 20 feet ahead of the third-place runner. Assuming that each runner maintains a constant speed throughout the race, by how many feet does the second-place runner beat the third-place runner?

Step-by-Step Solution

Answer

The second-place runner beat the third-place runner by $10.02$ feet.

Step 1. Forming equation for the speed of the three runners.

Let $v_1,v_2,v_3$ denote the speeds of the first, second, and third place runners, respectively.

Total length of race is mile, that is

$1\text{ mile }=5280\text{ feet }$.

Now, Let $t_1,t_2$ denote the times required for the first-place runner and second-place runner to finish the race. Then we have the system of equations :

1. For winner crosses the finish line

$v_1{t}_1=5280$

As we know that

$speed\times time= dis\tan ce$

2. For winner crosses the finish line 10 feet ahead of the second-place runner 

$v_2{t}_1=5270$

3. For winner crosses the finish line 20 feet ahead of the third-place runner 

$v_3{t}_1=5260$

4. For second-place runner crosses the finish line 

$v_2{t}_2=5280$

Step 2. Finding the required distance.

At time $t_2$, the third-place runner has gone a distance of $v_3{t}_2$ feet, so the distance d remaining is $5280-v_3{t}_2$.

Now

$$\begin{gathered} d=5280-v_3{t}_2 \\ =5280-v_3(t_1\times \frac{t_2}{t_1}) \\ =5280-v_3{t}_1\left(\frac{t_2}{t_1}\right) \\ =5280-\left(5260\right)\frac{t_2}{t_1} (v_3{t}_1=5260) \\ =5280-\left(5260\right)\frac{{\displaystyle \frac{5280}{v_2}}}{{\displaystyle \frac{5280}{v_1}}} (v_1{t}_1=5280 and v_2{t}_2=5280) \\ =5280-\left(5260\right)\frac{v_1}{v_2} \\ =5280-\left(5260\right)\frac{5280}{5270} (\frac{v_1{t}_1}{v_2{t}_1}=\frac{5280}{5270}) \\ =10.02 \end{gathered}$$

So the second-place runner beat the third-place runner by 10.02 feet.