Answer:
Hi! I'm pretty sure the question is: Two cities are 45 miles apart. Two trains, with speeds of 70 mph and 60 mph, leave the two towns 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?
I have had a similar question before, and I will answer yours. :)
Explanation:
T1 - the time the trains are moving to each other.
For time T1, the first train will go the distance of 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 into 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 meets 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 the beginning of the movement, it will be time when they met + time when they were moved 10 mi apart from each other
9/26 h+1/13h = 11/26 h = 0.42 h