Chapter 11.6, Problem 84AYU

To determine

To find: The distance by which the second place runner beat the third place runner.

Answer to Problem 84AYU

Solution:

10 feet

Explanation of Solution

Given:

• $\text{Length} \text{of} \text{the} \text{race} = 1 \text{mile} \text{or} 5280 \text{ft}$

• The first winner crosses the finish line 10 feet ahead of the second-place runner and 20 feet ahead of the third place runner.

Formula Used:

$\text{Speed} = \text{distance}/\text{time}.$

Calculation:

Let, $v_1, v_1, v_1$ be the speeds of 1^st, 2^nd and 3^rd runners respectively.

When the first runner finishes the distance of 5280 feet in time $t_1$, the second runner has covered 5270 feet and the 3^rd runner has covered 5260 feet.

In time $t_1$, the speeds of the three runners are

$v_1 = \frac{5280}{t_1}$   -----Eq(1)

$v_2 = \frac{5270}{t_1}$   -----Eq(2)

$v_3 = \frac{5260}{t_1}$   -----Eq(3)

Let $t_2$ be the time for the second runner to complete the race

$t_2 = \frac{5280}{v_2}$   -----Eq(4)

Substituting $v_2$ from Eq(2) in Eq(4)

$t_2 = \frac{5280}{5270}{t}_1$   -----Eq(5)

The distance travelled by third runner by the time $(t_2)$

$d_3 = v_3{t}_2$   -----Eq(6)

Substituting Eq (3) and Eq(5) in Eq (6) we get,

$d_3 = \frac{5260}{t_1} \frac{5280}{5270}{t}_1$

The distance by which the second runner is ahead of the third runner is

$\begin{array}{l}= 5280 - d_3\\ = 5280 \left(1 - \frac{5260}{5270}\right)\\ = 10 \text{feet}\end{array}$