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Subject: Uniform Motion Word Problem A Metra commuter train leaves Union Station in Chicago at 12 noon. Two hours later, an Amtrak train leaves on the same track, traveling at an average speed that is 50 miles per hour faster than the Metra train. At 3PM the Amtrak train is 10 miles behind the commuter train. How fast is each going?
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Hi Amy. Distance equals speed times time. So if the Metra train is travelling for 3 hours at speed "m" miles per hour, then it has gone 3m miles by 3 o'clock. Likewise, the Amtrak train has gone 1a miles by 3'oclock. At this time, the difference in their distances is 10 miles, so 3m - a = 10. You also know that the difference in their speeds is 50 miles per hour, so a - m = 50. This gives you two equations with two unknowns: 3m - a = 10 and a - m = 50. Solve one equation for one variable and substitute it into the second equation to get the value of the second variable, then use that value to determine the first variable. Hope this helps, Amy, There is another way to look at this problem that gives a second solution. The Metra leaves the station at noon and by 2pm has travelled some unknown distance. I am going to call this distance D miles, it is the distance between the two trains at 2 pm. The amtrack train starts at 50 miles per hour faster than the Metra and hence the distance between the trains begins to decrease. One hour later, at 3 pm, the Amtrack has travelled 50 miles and is only 10 miles behind the Metra. Thus D must have ben 60 miles. Hence from noon until 2 pm the metra travelled 60 miles. How fast was it going? Penny |
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