Fred begins walking toward John house at 3 mi/h. John leaves his house at the same time and walks toward Fred's house on the sam

Question

miskamm [114]

2 years ago

11

Fred begins walking toward John house at 3 mi/h. John leaves his house at the same time and walks toward Fred's house on the sam

e path at a rate of 2 mi/h. How long will it be before they meet if the distance between the houses is 4 miles?

Mathematics

1 answer:

Tcecarenko [31] · Answer 1 · 2022-04-19T20:42:36+03:00

Let us assume as at when they meet, the time elapsed is x hours.

Fred 3 mi/h
John 2 mi/h

Distance = speed * time

Fred = 3x
John = 2x

Houses are 4miles apart.
So: 3x + 2x = 4 5x = 4

x = 4/5 hour = 0.8 hour or 0.8*60 = 48 minutes

Time elapsed is 0.8 hour or 48 minutes.