SOLUTION: Train A leaves a station traveling at 48 kilometers per hour. Four hours later, train B leaves the station traveling in the same direction at 68 kilometers per hour. What equation

Question 94247: Train A leaves a station traveling at 48 kilometers per hour. Four hours later, train B leaves the station traveling in the same direction at 68 kilometers per hour. What equation can you use to find the time (t) in hours that it took train B to catch up with train A, and what is the value of t
Found 2 solutions by checkley71, edjones:
Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website!
48T=68(T-4)
48T=68T-272
68T-48T=272
20T=272
T=272/20
T=13.6 HOURS THE FAST TRAIN WILL CATCH THE SLOWER TRAIN.
PROOF
48*13.6=68(13.6-4)
652.8=68*9.6
652.8=652.8

Answer by edjones(7311) (Show Source):
You can put this solution on YOUR website!
d=distance t=time v=velocity
d=tv
Let t=time of 1st train
Let t-4=time of 2nd train
for the 1st train d=tv=48t
for the 2nd train d=tv=68(t-4)
they both travel the same distance therefore: 48t=68(t-4)
48t=68t-272
add 272 and -48t to both sides: 272=20t
t=13.6hrs
t-4=9.6hrs time it took train b to catch up to train a
EdJones