Question

Two athletes practice for a marathon by running back and forth on a 11-mile course. They start running simultaneously, one at a speed 2 miles faster than the other. How fast does each run if they meet 1 hr 6 minutes after starting? (The faster one is already returning at this point.) How far from the starting point do the runners meet?

Answer 1

Answer:

Faster runner speed: 11 mph

Slower runner speed: 9 mph

Distance from starting point: 9.9 miles

Step-by-step explanation:

Let the slower runner speed be: x

Faster runner speed: x+2

1 hour 6 minutes = 1.1 hours

Slower: 1.1(x) = 1.1x miles

Faster: 1.1(x+2) = 1.1x+2.2 miles

This means that the faster runner had gone 2.2 miles more than the slower runner.

1.1x + 2.2 > 11 miles (This is true as we know the faster runner is coming back from the 11 mile run)

1.1x > 11 - 2.2 miles

1.1x > 8.8 miles

x > 8.8 / 2.2 miles

x > 8 mph

1.1x < 11 miles (we know this is true as the slower runner has to have gone less than 11 miles as the faster runner met him on his way back)

x < 11 / 1.1 miles

x < 10 miles

8 < x < 10

Although x could be any number including decimal between 8 and 10, the obvious first answer would be 9. However we need to check if this answer is correct.

Slower runner distance to the end:

1.1(9) = 9.9 miles

11 - 9.9 = 1.1 miles

Faster runners distance from the end (when coming back)

1.1(9+2) = 1.1(11) = 12.1 miles

12.1 - 11 = 1.1 miles

The speeds 9 mph and 11 mph are now proven to be the right speeds. Using our equation before, we can figure out the distance from the starting point that they meet at. This is just how far the slower runner has gotten. So, 1.1(9) = 9.9 miles from the starting point

Answer 2

The runners meet at a distance of 8.8 miles from the starting point.

How to determine the time required for two runners to meet each other

According to statement, each runner must run back and forth on a 11-mile course. After supposing that each runner moves at constant speed, expressions for each runner are presented below:

Fast runner reaches the end

$(v+2)\cdot 1.1 = 11$ (1)

Fast runner comes back from the end

$x = \left(v+2\right)\cdot (1.1-t)-(v+2)\cdot t$ (2)

Slow runner meets fast runner

$x = v\cdot 1.1$ (3)

Where:

• $v$ - Speed of the slow runner, in miles per hour.
• $t$ - Comeback time of fast runner, in hours.
• $x$ - Point of the track where runners meet each other, in miles.

By (1), we obtain the speed of the slow runner:

$v = \frac{11}{1.1}-2$

$v = 8\,\frac{mi}{h}$

The position from the starting point is calculated by (3):

$x = (8\,mi)\cdot \left(1.1\,h\right)$

$x = 8.8\,mi$

The runners meet at a distance of 8.8 miles from the starting point. $\blacksquare$

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