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A Wild Ride

  1. Feb 1, 2008 #1
    1. The problem statement, all variables and given/known data

    A car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the x axis be parallel to the ground and the positive y axis point upward. In the time interval from t=0 to t=4 s, the trajectory of the car along a certain section of the track is given by

    where A is a positive dimensionless constant.

    The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20m/s. Find the maximum value of A allowed by these regulations.

    2. Relevant equations
    A = [tex]\sqrt{a_{x}^{2} + a_{y}^{2}}[/tex]

    3. The attempt at a solution
    well, I have separated this equation into 2 components of position, rx ry
    Rx = A(t)
    Ry= A(T[tex]^{3}[/tex] - 6T[tex]^{2}[/tex])

    took the derivative of each component to change R to V

    Vx = A (this is A because, A is a constant and i just treated this as i took the deriative of any constant next to a variable with power of 1, just kept the constant.)
    Vy = A(3T[tex]^{2}[/tex] - 12T)

    The magnitude of this vector V is
    V = [tex]\sqrt{ A^{2} + (3T^{2}-A12T)^{2}}[/tex]

    now.. my problem here is how can i find which maximum value of A whose speed doesnt pass 20, i HAVE thought of setting this equation to 20, but what about -20 velocity, since it asks for speed not velocity...this rollercoaster CAN go downward.

    any advice?
    Last edited: Feb 1, 2008
  2. jcsd
  3. Feb 3, 2008 #2
    Okay, so what is that you are supposed to do again? You are right, except your velocity isn't right (but your idea is!), so far as I can tell what you are supposed to do.
  4. Feb 3, 2008 #3
    im supposed to find what highest value of A can be put in so the rollercoaster doesnt speed past 20 m/s,

    homework was due , couldnt get it, answer was 1.666 ...how can you get this?
  5. Feb 3, 2008 #4
    Ah, okay. Whether or 20 is negative or positive would be irrelevant for the magnitude of the velocity.

    So let's continue with what you did:

    [tex]V = \sqrt{v_x^2 + v_y^2}[/tex]

    [tex]V = \sqrt{A^2+A^2(3t^2-12t)^2}[/tex]

    [tex]A = \frac{V}{\sqrt{1+(3t^2-12t)^2}}[/tex]

    So now we see that the amplitude will depend on time because the velocity is defined to be less than 20m/s. We also want to maximum, so let's take the first derivative of A with respect to time.

    [tex]\frac{dA}{dt} = 0 = -\frac{1}{2} \frac{V}{\sqrt{1+(3t^2-12t)^2}}*2(3t^2-12t)(6t-12)[/tex]

    This equation looks sort of beefy, and because I have done enough algebra to last a lifetime I put it into mathematica.

    Mathematica, a math program, tells me (after plugging in 20 for V) that the time derivative is zero, i.e. the maxima occur at 0,2,4. The zero and 4 times are out of our control, and we don't really care about them. So let's go back to A(t) and plug in 2 for t.

    A(2) = 1.661
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