SOLUTION: A Car A left City X, traveling at an average velocity of 50 miles per hour. Three hours later a Car B left City X traveling on the same road at 60 miles per hour. When will the Car

Question 671618: A Car A left City X, traveling at an average velocity of 50 miles per hour. Three hours later a Car B left City X traveling on the same road at 60 miles per hour. When will the Car B catch up with the Car A? How far will each car have traveled?
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Use d = rt
For car A we have
(1) dA = 50t
For car B we have
(2) dB = 60(t-3) because car B left 3 hrs after car A
When they meet they both travelled the same distance say dF, then we have
(3) dF = dA = dB
Setting (1) equal to (2) we get
(4) 50t = 60(t-3) or
(5) 50t = 60t - 180 or
(6) 10t = 180 or
(7) t = 18 hrs
The distance travelled by car A in 18 hrs is
(8) dF = 50*18 or
(9) dF = 900 mi
To check this, let's see how far car B travelled. Using t = 18 in (2) we get
(10) dB = 60(18-3) or
(11) dB = 60*15 or
(12) dB = 900 mi
Answer: Car B will catch up with car A 18 hours after car A left city X at a distance of 900 miles.

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