SOLUTION: a group of people started driving at 6 mph. one of the cars didnt move until its started driving 10 mph. the car caught up to the others in 5 minutes.
question:what is the dis
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question:what is the dis
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Question 55838This question is from textbook
: a group of people started driving at 6 mph. one of the cars didnt move until its started driving 10 mph. the car caught up to the others in 5 minutes.
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? This question is from textbook
You can put this solution on YOUR website! 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
:
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
:
1.24 mi traveled when car catches up with the group
