Question 102340: 1. A heavy freight train leaves New York for Chicago at 7 A.M., going 40 miles an hour. Another train follows at 7:30 A.M., going 50 miles an hour. At what time does the second train overtake the first, and how far from New York?
4. A boat left X at 8 A.M., going 30 miles an hour. At 9:30 A.M. it had engine trouble that delayed it for 1 hour. Then it continued at the reduced speed of 10 miles per hour. A second boat left X at 10:30 A.M. and traveled 40 miles per hour. At what time did the second boat overtake the first boat. How far from X was the meeting point?
I don't even know how to set this problem up. Thanks for the help!
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! If you think carefully you can work both of these problems in your head. Let's try that first.
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For the train problem, the first train has a half hour head start. At 40 mph this means that
the first train has gone 20 miles before the second train leaves. The second train is going
50 mph. That means that it is going 10 mph hour faster than the first train. At 10 mph faster
it will need 2 hours to make up the 20 miles it is behind. Therefore, 2 hours after the
second train departs it catches up to the first train. Since the second train left at 7:30 a.m.
it catches up to the first train at 9:30 a.m. And at 50 mph, the second train travels
100 miles from New York in the 2 hours it takes to catch up to the first train.
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Now for the equation form. You could set it up this way ... let t equal the amount of
time that the first train is moving. The distance it will move in t hours at 40 miles per
hour is 40t. The amount of time that the second train travels is a half hour less than t
which is (t - 1/2) hours. At 50 mph the second train travels a distance of 50*(t - 1/2).
Since the two trains will be an equal distance from New York City when they catch up,
you can say that the distance of one is equal to the distance of the other. So the equation
is:
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40t = 50(t - 1/2)
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Multiply out the right side and the equation becomes:
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40t = 50t - 25
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Get rid of the 50t on the right side by subtracting 50t from both sides to get:
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-10t = -25
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Solve for t by dividing both sides by -10 and the answer is:
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t = 2.5 hours
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Since we said t was the time that the first train traveled, and since the first train left
at 7:00 a.m. then 2.5 hours after 7:00 a.m. is 9:30 a.m. The first train is going at
a rate of 40 mph, so in 2.5 hours it goes 40 times 2.5 = 100 miles. These are the same
answers that we got above by thinking our way through the problem.
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The boat problem can be solved the same way. Think this way ... the boat travels at 30 mph
for the first hour and a half. At 30 mph for 1.5 hours it goes 45 miles (30 miles in the
first hour plus 15 miles in the next half hour). Then it breaks down and sits there for
an hour. So in 2.5 hours (10:30 a.m.) it is 45 miles ahead of the second boat. The second boat
then departs going 40 mph while the first boat is now limping along at 10 mph. So the
second boat is making up mileage at the rate of 30 mph. At this "make-up" rate how long
does it take the second boat to make up for the 45 mile head start? In the first hour the
second boat travels it makes up 30 miles. In the next half hour it makes up the other 15
miles. So, the second boat catches up in 1.5 hours. Since it departed at 10:30 a.m. and
travels for 1.5 hours, the second boat catches the first boat at noon. In the 1.5 hours
the second boat travels at 40 mph, it travels 40 times 1.5 which is 60 miles. The first
boat traveled 45 miles up to 10:30 a.m. In the next 1.5 hours at the rate of 10 mph it
goes 10 times 1.5 = 15 miles. So the total distance it traveled by noon is also 60 miles.
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The equation form of this starts with the first boat. It goes 30 mph times 1.5 hours and
then it breaks down. It goes 0 mph for 1 hour. Then it goes 10 mph for an unknown time t.
The distance that it goes is, therefore: 30*1.5 + 0*1 + 10t
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This simplifies to: 45 + 10t
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The second boat goes at 45 mph for time t after 10:30. So it goes 45*t and that must equal the
distance traveled by the first boat. Setting the two distances equal results in the equation:
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45 + 10t = 40t
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Subtract 10t from both sides and you have:
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45 = 30t
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Solve for t by dividing both sides by 30 to get:
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1.5 = t
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Since t was the time traveled by the second boat ... that departed at 10:30 a.m. the second
boat travels until noon at which time it has caught up. And again, at 40 mph, in the 1.5 hours
it travels the second boat goes 40 times 1.5 miles or 60 miles.
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Hope this helps you to understand the problem a little better.
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