September 21, 2002
Kevin Eger
Franklin & Marshall College
Lancaster, PA 17604
F.A.R.
2468 Gowa Way
Shoe Chew City, PU 11248
Dear Mr. G. Olson Overby-Fitzpatrick,
My name is Kevin Eger, and I am a Calculus student for Dr. Crannell. I was extremely excited when I received your letter. I found your dilemma to be very intriguing when you stated that you wanted to know the speed of the train for the last leg of your trip to Sugar. I am sorry to inform you this, but it is true that your train was in fact traveling at 75 M.P.H. for the last 20 miles of the trip. Let me show you what formula I used and how I came to my final conclusions.
First I laid out all of the facts of the case in front of me. We know it takes 60 minutes to travel to Sugar, which is 60 miles away. The next thing we know is that when Pete traveled to Sugar that day, the train traveled at 50 M.P.H. for the first 20 miles of the trip. Also we recognize that for the next 20 miles, the train traveled at 60 M.P.H. So what we want to know is, how fast was the train traveling for the last 20 miles of the trip?
To figure out how fast the train was traveling for the last leg of the trip, we first have to calculate how long the first two legs of the trip took. To figure out the time for each leg of the trip, I used the formula
In this equation, distance represents how far the train traveled, and rate represents the speed of the train. This is a basic formula that can be derived from many resources, such as a Calculus book. In this formula, you must assume that the rate of speed is constant. If the train slows down or changes speed through one of the legs, then the calculations could be wrong. So when using this formula I assumed that the rate of speed in each leg was at a constant rate of speed.
I took 20 miles (the distance of travel for the first leg) and divided it by 50 M.P.H. (the speed of the train during the first 20 miles.) After dividing the two numbers, I received an answer of .4. This means that it took Pete .4 hours to travel 20 miles. The next step I took was to multiply .4 by 60, because there are 60 minutes in an hour. When I multipied .4 by 60, I got an answer of 24, which stands for 24 minutes. Now we know that the first leg of the trip took 24 minutes.
I used the same procedure to calculate the second leg of the trip. I took 20 miles (distance of travel for the second leg) and I divided it by 60 M.P.H. (the speed of the train during the second 20 miles.) After dividing the two numbers, I got an answer of .333. This means that it took Pete .333 hours to travel 20 miles. I took that number and multiplied it by 60 to receive an answer of 20, which stands for 20 minutes. Now we know that the second leg of the trip took 20 minutes.
So far we know that the first and second legs of the trip took 44 minutes. This means that the third leg of the trip must take exactly 16 minutes, because you said that you had witnesses in the station that can swear the train was on time. If you add 44 minutes and 16 minutes, you get 60 minutes, which is how long it took to make the trip.
To figure out the speed of the train for the third leg, I simply plugged in a number for the speed. The first number I plugged in for the speed was 70 M.P.H. If you take 20 miles (distance of travel for the third leg) and divide it by 70 M.P.H. (the speed we are assuming the train was traveling), you get .286. This means that it took Pete .286 hours to travel 20 miles. Next you take .286 and times it by 60 and you get an answer of 17.143 minutes. We know that the last leg should have been completed in 16 minutes. So if the train were traveling at 70 M.P.H. the train would have been 1.143 minutes late.
If you repeat this last step, but plug in 75 M.P.H. for the speed, you will see that the train does in fact come in at exactly 60 minutes. Just to make sure that my calculations were indeed correct, I showed my work to Adam Mistick and John Pittenger, who are fellow Calculus students. They agreed that the train was traveling at 75 M.P.H. I also questioned Dr. Crannell, who was very helpful, and she too agreed that the train was traveling at 75 M.P.H.
The reason that your calculations were not correct is because the time of each leg was not constant. If each leg took 20 minutes, then you could have taken the average speeds of 50, 60, and 70, and received an average speed of 60 M.P.H and arrived to the station on time. Since the legs took 24, 20, and 16 minutes each, the train must have traveled faster for each leg of the trip to make it in 60 minutes.
I hope that my calculations were easy enough for you to understand and I wish you the best of luck on your court date.
Best Wishes,
Kevin Eger
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