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Math Help - How do I set up this problem?

  1. #1
    Member Ranger SVO's Avatar
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    How do I set up this problem?

    A car travels at 60mph, a truck travels 55mph in the opposite direction. The vehicles pass in .34 seconds. How long is the truck?

    How do I set this problem up??
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: How do I set up this problem?

    (60/3600+55/3600)*0.34*1609.344 = 17.479 meters
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    Re: How do I set up this problem?

    60 mph = 88 ft/sec

    55 mph = (242/3) ft/sec

    rate of closure = 88 + (242/3) = (506/3) ft/sec

    (rate)(time) = distance

    (506/3 ft/sec)(34/100 sec) ... approx 57.35 ft
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  4. #4
    Member Ranger SVO's Avatar
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    Re: How do I set up this problem?

    Thanks to both of you, it is appreciated.
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  5. #5
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    Re: How do I set up this problem?

    The solutions above assume that the time it takes for the vehicles to pass is measured from the moment when their front bumpers align to the moment when the front bumper of the car aligns with the back of the truck. During this time, the car travels the whole truck's length relative to the truck. This is why the rate of closure times 0.34 s presumable equals the length of the truck. However, at the end of this time, the car and the truck still overlap. If we measure the passing time as the time when the vehicles overlap, then the rate of closure times 0.34 s equals the combined lengths of the vehicles.
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  6. #6
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    Re: How do I set up this problem?

    Hello, Ranger SVO!

    A car travels at 60mph, a truck travels 55mph in the opposite direction.
    The vehicles pass in 0.34 seconds. .How long is the truck?
    Is that a misprint?

    If this is conventional "passing vehicles", there is no unique solution.
    . . We would need the length of the car.

    Normally, the passing procedure begins when their front bumpers are even
    . . and ends when their rear bumpers are even.

    Code:
              * - - *
              | car | →
              * - - *
    Start           * - - - - *
                  ← |  truck  |
                    * - - - - *
    
    
                      * - - *
                      | car | →
     End              * - - *
            * - - - - *
          ← |  truck  |
            * - - - - *
    Let C = length of the car.
    Let T = length of the truck.

    They have a combined speed of 115 mph.

    It is as if the car has stopped
    . . and the truck is passing at 115 mph.

    The truck travels 5280(C+T) feet at \tfrac{506}{3} ft/sec
    . . and takes 0.34 seconds.

    We have: . 5280(C+T) \:=\:\tfrac{506}{3}(0.34) \quad\Rightarrow\quad C+T \:=\:0.03258333 feet.

    The combined lengths of both vehicles is less than a half-inch!
    And we still don't know the value of T.
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  7. #7
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    Re: How do I set up this problem?

    Quote Originally Posted by Soroban View Post
    The truck travels 5280(C+T) feet at \tfrac{506}{3} ft/sec
    . . and takes 0.34 seconds.
    Shouldn't this say, "The truck travels (C + T) feet at \tfrac{506}{3} ft/sec and takes 0.34 seconds"?
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  8. #8
    Member Ranger SVO's Avatar
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    Re: How do I set up this problem?

    Skeeters answer matches the answer key. When converted so does MaxJaspers answer.

    Answer keys are worthless if you can not work the problem, this problem is an example UIL calculator competition problem.
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  9. #9
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    Re: How do I set up this problem?

    Quote Originally Posted by Ranger SVO View Post
    Skeeters answer matches the answer key. When converted so does MaxJaspers answer.

    Answer keys are worthless if you can not work the problem, this problem is an example UIL calculator competition problem.


    The answer key is wrong as shown by emakarov
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