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Homework Help: Determining Time When Comparing Two Equations

  1. Jan 7, 2004 #1

    JDK

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    Hello all,

    The last time I used these forums for help I got awesome responses. And here I am again needing help. Thought I'd just mosey on over and pop up my question I got. Here it is:

    (2) A car and motorcycle start from rest at the same time form rest on a straight track, but the motorcycle is 25.0 m behind the car. The car accelerates at a uniform rate of 3.70 m/s2 and the motorcycle accelerates at a uniform rate of 4.40 m/s2.
    (a) How much time elapses before the motorcycle overtakes the car?

    I know that I need this formula --> d = Vi(t) + 1/2(a)(t^2)
    And I know that it should be set up so that the two distance equations (one for the car and one for the motorcycle) are set equal to each other, so I can further solve the junctioned equation for (t). But, I seem to be having some trouble. I need to know how to go about solving this problem. Any help is very much appreciated! Thanks a lot.

    - JDK
     
  2. jcsd
  3. Jan 7, 2004 #2

    HallsofIvy

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    Science Advisor

    The formula "d = Vi(t) + 1/2(a)(t2)" isn't quite correct. It should be d = Vi(t) + 1/2(a)(t2)+ di where di is the initial distance from some reference point.

    In this case, if you take the point from which the motorcycle starts as the reference point, then di= 0 for the motorcycle and di= 25 for the car (since the car starts 25 meters ahead of the motorcycle).
    Of course, Vi= 0 for both motorcycle and car.

    For the car, a= 3.7 m/s2 so d= (3.7/2)t2+ 25= 1.85 t2+ 25.
    For the motorcycle a= 4.4 m/s2 so d= (4.4/2)t2+0= 2.2 t2.

    The motorcycle will overtake the car when the two d s are the same:
    when 1.85 t2+ 25= 2.2 t2. That's the equation you want to solve.
     
  4. Jan 8, 2004 #3

    JDK

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    Thank you. Now I understand what I need to do. My problem was not knowing the 'Initial Distance Variable'.

    ...

    Vi(t) + ½(Ac)(t²) + Dic = Vi(t) + ½(Am)(t²) + Dim
    (0)(t) + ½(3.70 m/s²)(t²) + 25m = (0)(t) + ½(4.40 m/s²)(t²) + 0m
    ½ (3.70 m/s²)(t²) + 25m = ½(4.40 m/s²)(t²)
    (1.85 m/s²)t² + 25m = (2.2 m/s²)t²

    ... excuse my lack of knowledge in this area but next am I supposed to square each side of the equation or divide the right side by (1.85)? Or is it something completely different. I seriously must be having an off day... *embarassment* This usually doesn't happen... I've tried a few things on it but I have no idea what the right answer is... that's my problem right now...
     
  5. Jan 8, 2004 #4

    Doc Al

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    Staff: Mentor

    Combine the two terms that contain t2. (Subtract!)
     
  6. Jan 8, 2004 #5

    JDK

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    *smacks self promptly*

    Pheww... I better do a brain check... anyways, so this is what it should look like then - correct? (My symbols are showing up all googled... I'll change that...)

    (1.85 m/s²)t² + 25m = (2.2 m/s²)t²
    25m = (2.2 m/s²)t² - (1.85 m/s²)t²
    25m = (0.35 m/s²)t²
    25m / 0.35 m/s² = t²
    71.4 s² = t²
    (sqroot)71.4 s² = t
    8.4 s = t

    Thanks by the way! :)
     
    Last edited: Jan 8, 2004
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