A Wild Ride - roller coaster physics

[rsala]

Homework Statement

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
http://img526.imageshack.us/img526/719/renderxg2.gif

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.

Homework Equations

A = \sqrt{a_{x}^{2} + a_{y}^{2}}

The Attempt at a Solution

well, I have separated this equation into 2 components of position, rx ry
Rx = A(t)
Ry= A(T^{3} - 6T^{2})

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^{2} - 12T)

The magnitude of this vector V is
V = \sqrt{ A^{2} + (3T^{2}-A12T)^{2}}now.. my problem here is how can i find which maximum value of A whose speed doesn't 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?

 

[Mindscrape]

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.

 

[rsala]

im supposed to find what highest value of A can be put in so the rollercoaster doesn't speed past 20 m/s,

homework was due , couldn't get it, answer was 1.666 ...how can you get this?

 

[Mindscrape]

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:

V = \sqrt{v_x^2 + v_y^2}

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

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

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.

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

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