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A really annoying speedboat.

  1. Jul 30, 2007 #1

    A speedboat moving at 30.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.50 m/s^2 by reducing the throttle.
    a. How long does it take the boat to reach the buoy?
    b. What is the belocity of the boat when it reaches the buoy?

    Equations given:
    V[final] = V[initial] + AT (for constant acc.)
    X[final] = X[initial] + .5(V[initial] + V[final])T
    X[final] = X[initial] + (V[initial])T + .5AT^2

    My pitiful attempt:
    For a. : I used the third equation, assuming that my initial position would be zero, as the point from which the deceleration began, and my final position would be 100. The two results for the quadratic equation were 12.612 and 4.531.

    That's where I was stuck.

    I then tried b. : Utilizing the second equation, I got V^2 = 200 (m^2 - m)/(s^2)

    ...this just doesn't seem right, and I would really appreciate some help.
  2. jcsd
  3. Jul 30, 2007 #2


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    Are you sure you copied those equations correctly,
    The ussual form of the equations of motion is:
    v = u+at
    s = ut + 0.5 at^2
    v^2 = u^2 + 2as

    where u=initial veleocity, v=final velocity, s=distance
    Just look at what numbers you know and pick the appropriate equation.
  4. Jul 30, 2007 #3
    Gosh, I wish my physics book explained it as simply as that. The equations you gave me are probably there, somewhere, in my chapter, but in more superfluous terms (as I haven't encountered the equations you presented).

    I'll try the problem again. :)
  5. Jul 30, 2007 #4


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    The equations are given properly typeset in the introduction to this thread.

    You intro physics textbook should explain how to derive them from the definitions of velocity and acceleration, but it's worth memorising these 3.

    Another tip - check the units when rearranging equations or calculating an answer. Remember you can only add the same units but you can multiply/divide different ones. eg you can divide velocity by time to get acceleration but if you find yourself trying to add a velocity to a distance you have done soemthing wrong.
  6. Jul 30, 2007 #5

    The equations you gave me are the same ones my book provided, only my book lengthened it by plugging in how to the get the total displacement and whatnot.

    However, I was wondering if my train of thought is correct by assuming the displacement to be 100. Would it matter if I believed X[initial] to be 0 or 100, and X[final] to be 100 or 0, respectively? It's just a matter of a negative sign, but, would my answer be correct no matter which two numbers the two variables stood for?
  7. Jul 31, 2007 #6


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    Your two times seems to be corrrect (I get the same values except the second one I get 4.54 s).

    Why do you get two answers?

    The same reason why you will get two answers if you were to calculate when a ball that is thrown upwards will pass the same point above the thrower (if the point is below the maximum height reached).
    Last edited: Jul 31, 2007
  8. Jul 31, 2007 #7
    Its probably worth stating just how important these equations are in mechanics. After the obvious F=ma, these are probably the most fundamental and widespread equations you will come across if you continue work on motion (with constant acceleration)
  9. Jul 31, 2007 #8
    Heh, yes, I'm beginning to see that. (I've started to refer to them as The Trifecta).

    I think I've got the answer, by first figuring out (b). Using V^2 = U^2 + 2AS, I found the final velocity to be about 14.142 (I will round to the proper amount of sig figs later) or the square root of 200. After I have the final velocity, I can plug everything into the equation V = U + AT to get T = 4.531 seconds. I don't exactly know how I got another possible answer (not unusual or negative) for time when using the third equation, but I'll look over the equation again.

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