Two trains, Train A and Train B, simultaneously depart Station A and Station B. Station A and Station B are 252.5 miles apart from each other. Train A is moving at 124.7mph towards Station B, and Train B is moving at 253.5mph towards Station A. If both trains departed at 10:00AM and it is now 10:08, how much longer until both trains pass each other?
+817 Train A
124.7mph (124.7/60*8)=16.162mi
124.7mph/60mins*8(mins)
Train B
253mph (253/60*8)=33.7mi
33.7+16.16=49.89mi
252.5-49.89=203.61mi (left to travel)
124.7+252.5=377.2mph (combined speed)
203.61mi / 377.2mph =.539hrs (time to come together)
so 32 mins, so 10:08+32mins= 10:40
+128707 Very nice, Landry......here's another approach
The combined rates = 377.2
And the total distance to cover = 252.5 miles .......so.....
Distance / Rate = Total Time for the trip in hours ..... and we have
252.5 / 377.2 = about .668 hrs
And in minutes this is about .668 x 60 = 40 minutes
And....since they have already traveled 8 minutes, they have 32 more to travel .........so they will meet at 10:40..........as Landry found
+128707 I just realized that the combined rate we gave was in error....it should have been 378.2....so.....
The total time they take from the time they start until the time they meet [in minutes] is given by
[252.5/378.2] * 60 = about 40.058 minutes.....and since they have traveled 8 minutes already......they will meet in 32.058 minutes
This is still close enough to our previous answer of ≈ 32 minutes
