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Thread: motion word problem: At 9:45 a bus left Point Reyes, CA, going north at 40 mph....

  1. #11
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    Quote Originally Posted by allegansveritatem View Post
    Well, I don't know why I didn't see that I was mixing apples and oranges with the term with the ratio....I can give some account of my reasoning however: I had just dealt with another problem similar to this The other poblem stated:a bus and a car travel the same route both leaving from the same point. The bus goes 72kmh, the car goes 80kmh. If the bus leaves one hour before the car, how long will it take the car to catch up to the bus. In solving this problem I used the ratio method and it worked! Here is the equation I devised: 72t = 80(t-.9). The solution is : 9. The back of the book confirmed my result. I don't know how the author solved this problem but at least in this instance the ratio method hit the target. Maybe this was just coincidence. I haven't had time to test it. Obviously it doesn't work in the present porblem.

    I think my reasoning in this case was something like: the ratio of speeds indicates the lag between the speeds and this indication can be used somehow to sub for a time value. I don't know. I guess what killed me was the memory of how well this method worked at least once before in a similar problem.

    I like ratios. I like to find ways to use them. And obviously, to abuse them as well.
    The traditional algebraic way to solve this one would be to define t as the time after the car leaves, so that the bus takes t+1 hours. Then 72(t + 1) = 80t. This is equivalent to your equation. I think it was just luck (and the fact that the difference is 1 hour) that your work gave the right answer.

    For non-algebraic methods, what Denis did is what I would do. But you are probably in a class where you are expected to learn algebra, not general problem-solving skills, as valuable as the latter are.

  2. #12
    Quote Originally Posted by Denis View Post
    Quickest way for this problem is pretend the 2 buses are leaving at
    same time, distance 80 miles apart (the 1st 2 hours bus#1 travelled).

    Left to travel is 230-80 = 150 miles.
    At combined speed of 75mph, so another 2 hours.
    Game over!
    Yes...short and sweet.

  3. #13
    Quote Originally Posted by Dr.Peterson View Post

    For non-algebraic methods, what Denis did is what I would do. But you are probably in a class where you are expected to learn algebra, not general problem-solving skills, as valuable as the latter are.
    I am retired, not in any class. I took up algebra to keep my mind active. When I studied this subject in school I didn't have time to give it any real thought. I didn't experiment with it, didn't try to come up with my own solutions...because there was just no time for that. Too much pressure from too many sides. Now I have the time. I have no teacher...but with some help from people on this forum I think I will be OK. I have been hunting around online for the instreuctor's solutions manual for the old algebra book I have (Blitzer, 1994). So far no luck.

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