196k views
0 votes
Two athletes practice for a marathon by running back and forth on an 11-mile course. They start running simultaneously, one at a speed 2 mph faster than the other's speed. How fast does each run if they meet 1 hour 6 minutes after starting?

1 Answer

1 vote

Final answer:

The slower athlete runs at a speed of 4 mph, and the faster athlete runs at a speed of 6 mph. This was determined using the combined distance covered by the athletes and the time taken to meet.

Step-by-step explanation:

The question asks us to find the speeds of two athletes running back and forth on an 11-mile course, starting simultaneously, and meeting each other after 1 hour and 6 minutes. Let's denote the speed of the slower athlete as x mph. This means the faster athlete runs at x + 2 mph. To determine their speeds, we'll use the fact that their combined distance covered will be 11 miles when they meet.

Since they meet after 1 hour and 6 minutes, we convert the minutes into hours to make the calculations easier. The time in hours is 1 + 6/60, which simplifies to 1.1 hours. So, the distance the slower runner covers is x × 1.1 miles, and the distance the faster runner covers is (x + 2) × 1.1 miles.

The sum of these distances is equal to 11 miles, which gives us the equation:

1.1x + 1.1(x + 2) = 11

Solving for x we get:

2.2x + 2.2 = 11

2.2x = 11 - 2.2

2.2x = 8.8

x = 8.8 / 2.2

x = 4

Therefore, the slower athlete runs at 4 mph, and the faster athlete runs at 4 + 2 = 6 mph.

User Brechmos
by
8.2k points