Question

Two cities are 45 miles apart. Two trains, with speeds of 70 mph and 60 mph, leave the two cities at the same time so that one is catching up to the other. How long after the trains leave will the distance between them be 10 miles? How long after the trains leave will they be 10 miles apart for the first time? What about the second? PLEASE HELP ASAP I WILL MARK BRAINLIEST

Answer 1

The distance between the trains is changing at the rate of (70 -60) = 10 mph. They will be in the same place (45 mi)/(10 mi/h) = 4.5 hours after they leave their respective cities. They will be 10 miles apart both 1 hour before that time and one hour after that time.

The trains will be 10 miles apart 3.5 hours after leaving.

They will be 10 miles apart the second time 5.5 hours after leaving.

Answer 2

so basically, one train has a head start of 45 miles

we assume that the slower train has the head start because if the faster train had the head start, then faster train would just get farther and farther away from the slower train (distance≥45mi for all time)

so

set a 0 point

lets say that at t=0, fast train has traveled 0 and slow train has traveled 45

distance=speed times time+headstart

so dist_fast=70*t+0=70t

dist_slow=60*t+45=60t+45

when will the distance between them be 10?

there will be a time when slower train will be in front and when fast train will be in front

when slow train is in front, dist_slow-dist_fast=10 or

60t+45-70t=10

solve for t

60t+45-70t=10

-10t+45=10

minus 45 both sides

-10t=-35

divide both sides by -10

t=-3.5 hrs

when fast train is ahead, dist_fast-dist_slow=10

70t-(60t+45)=10

70t-60t-45=10

10t-45=10

add 45 to both sides

10t=55

divide both sides by 10

t=5.5 hrs

so after 3.5hrs is the first time they are 10 miles apart

after 5.5 hours is the 2nd time they are 10 miles apart