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A Wild Ride

  1. Feb 4, 2005 #1
    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

    (look at attachment)

    where [tex]A[/tex] is a positive dimensionless constant.

    I have two questions that i dont get:

    1.) Derive a general expression for the speed [tex]v[/tex] of the car.
    Express your answer in meters per second in terms of [tex]A[/tex] and [tex]T[/tex].

    if i get the derivate of r, respect to time, that would get me the velocity right? so... 3at^2-12t+a

    2.) 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 [tex]A[/tex] allowed by these regulations.

    V = V(0) + at
    20 = 0 + a(4)
    a= 5

    did i get any of this right? if not, please give me a hand

    Attached Files:

  2. jcsd
  3. Feb 4, 2005 #2
    in general,, if [itex]\vec{r}[/itex] is a vector equation s.t.

    [tex] \vec{r} (t) = f(t) \vec{x} + g(t) \vec{y} + h(t) \vec{z} [/tex]

    then the derivative of [itex] \vec{r} (t) [/itex] wrt t is given by,

    [tex]\frac{d}{dt}\vec{r} (t) = [ \frac{d}{dt} f(t) ] \vec{x} + [ \frac{d}{dt} g(t) ] \vec{x} + [ \frac{d}{dt} h(t) ] \vec{z} [/tex]

    where [itex] \vec{x}, \ \vec{y}, \ \vec{z} [/itex] are the unit vectors along the x, y, z, axis respectively.

    if [itex] \vec{r}(t) [/itex] is a vector vlalued function for the position of a particle, then [itex] \frac{d}{dt} \vec{r} (t) [/itex] gives you a vector function for the velocity. To find the speed, you just need to find the magnitude of the velocity function. This is done the same way you find the magnitude of a vector only now the componets are functions.
    ( This is usually taught in multivariable calculus )
    Last edited: Feb 4, 2005
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