Question 55838
A group of people started driving at 6 mph. One of the cars didn't move until it's started driving 10 mph. The car caught up to the others in 5 minutes.
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Question:  What is the distance of the group of people while the car that was behind started to catch up? What would be the expression?
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Since we are dealing in minutes, change the speed to miles/min:
6 mph = 6/60 = .1 mi/min
10 mph = 10/60 = .167 mi/min 
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When the faster car catches up with the group, they will have traveled the same distance. Dist = speed * time; 
    t = time the 1st group has traveled; 
(t-5) = time the late car has traveled;
:
.1t = .167(t-5)
.1t = .167t - .833
.1t - .167t = .833
-.067t = -.833
 t = -.833/-.067
 t = 12.438 minutes when the car and the group are together
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Group Distance = .1 * 12.438 = 1.24 mi at that time
 Car Distance  = .167 * (12.438-5) = 1.24 mi also traveling for only 7.438 min
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1.24 mi traveled when car catches up with the group