SOLUTION: Shelia leaves on a long trip driving at a steady rate of 30mph. Her sister Allison leaves from the same location traveling to the same destination 2 hours later. She drives at a s
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Question 81431: Shelia leaves on a long trip driving at a steady rate of 30mph. Her sister Allison leaves from the same location traveling to the same destination 2 hours later. She drives at a steady rate of 60mph. How long after Allison leaves home does she catch up to Shelia? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Shelia leaves on a long trip driving at a steady rate of 30mph. Her sister Allison leaves from the same location traveling to the same destination 2 hours later. She drives at a steady rate of 60mph. How long after Allison leaves home does she catch up to Shelia?
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Let t = time on the road when A catches up with S
Then (t+2) = time S is on the road, when A catches up
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We know that when A catches S, they will have traveled the same distance.
We can write a distance equation from that fact: Distance = speed * time
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A's dist = S's distance
60t = 30(t+2)
60t = 30t + 60
60t - 30t = 60
30t = 60
t = 60/30
t = 2 hrs for A to catch up with S
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Check our solution using the distance (S's time; 2 + 2 = 4 hrs)
30 * 4 = 120
60 * 2 = 120
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Did this make sense to you? Any question?
