A Car in a Roller Coaster Moves Along a Track That Consists of a Sequence of Ups and Downs

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A Wild Ride - roller coaster physics

• Thread starter rsala
• Start date Feb 1, 2008

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 [Broken]

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 = [tex]\sqrt{a_{x}^{2} + a_{y}^{2}}[/tex]

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 by a moderator: May 3, 2017

Answers and Replies

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.

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?

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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A Car in a Roller Coaster Moves Along a Track That Consists of a Sequence of Ups and Downs

Source: https://www.physicsforums.com/threads/a-wild-ride-roller-coaster-physics.212529/

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