Question

At 11:30 AM. two people leave their homes that are 18 miles apart and begin walking toward each other. If one person walks at a rate that is 2 mph faster than the other andthey meet after 1.5 hours, how fast was each person walking?

Answer 1

`5\text{ mph and 7 mph}`

Step-by-step explanation:

Let the speed of the first person be x mph and the speed of the second person be y mph

Since one person's speed is faster than the other, we can have:

`y\text{ = \lparen x + 2\rparen mph}`

Mathematically, distance equals the product of speed and time

The time traveled by each person is 1.5 hours

The distance traveled by each of them is:

`\begin{gathered} 1.5\text{ }*\text{ x = 1.5x miles} \\ 1.5(x+2)\text{ = \lparen1.5x + 3\rparen miles} \end{gathered}`

The sum of the two equals 18 miles

Thus:

`\begin{gathered} 1.5x\text{ + 1.5x + 3 = 3x + 3 = 18} \\ 3x\text{ = 18-3} \\ 3x\text{ = 15} \\ x\text{ = }(15)/(3) \\ x\text{ = 5 mph} \end{gathered}`

Recall;

`y\text{ = x + 2 = 5 + 2 = 7 mph}`

This means that the first person was walking 5 mph while the second was walking 7 mph