Published 11.05.2018 on subject Mathematics by Guest

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 they be 10 miles apart for the first time? How long after the trains leave will they be 10 miles apart for the second time?

Answered by Guest

T1  - the time the trains are moving to each other.
For time t1, first train will go the distance 70*t.
The second train, for time t will go the distance 60*t.
Both for time t1  - 70t+60(t1)=(70+60)*(t1).
45 - (70+60)*t is the distance between trains that left after t hours.
For the first time they will be 10 miles apart when they are moving to each other
45 - (70+60)*t = 10
45-10=(70+60)*t
35=130*t
t=35/130 h =7/26 h≈ 0.27 h

The train meet when distance=0
45 - (70+60)*t=0
45=130t
t=45/130 h = 9/26 h ≈0.35 h when trains met.

Trains began to move apart from each other after 9/26 h moving,
After t2 time trains will be (70+60)*(t2)=130*(t2) mi apart from each other.

130*(t2)=10
t2=1/13h≈0.08h after they met, they are apart from each other 10 mi.

so after beginning of the movement it will be 
time when they met + time when they are moved  10 mi apart from each other
9/26 h+1/13h = 11/26 h≈ 0.42 h