Solving the Roller Coaster Speed Problem: A Max Speed of 20

[theowne]

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

r = At(xhat) + A(t^3 - 6t^2)(y hat)

where A is a positive dimensionless constant.

A) Derive a general expression for the speed v of the car.

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

Homework Equations

r = At(xhat) + A(t^3 - 6t^2)(y hat)

The Attempt at a Solution

I think I can do the first part properly. I found the derivative of the xhat equation and the y hat equation, and then found the magnitude of speed using c^2 = a^2 + b^2 of the derivatives, which comes out to

v = sqrt [A^2 + (A (3t^2 - 12t))^2]

or simplified

v = A * sqrt[1 + 9t^2 * (t-4)^2]

THe program tells me this answer is correct.

I don't know how the do the last part of the equation at all. A is an unknown, but t is a variable as well...I tried using a hint in the program but it just tels me the answer should be vmax = A * sqrt [145]

I thought maybe, you find the derivative of v(x) and check when it's zero, then you know it's maximum velocity. But in the answer given, the derivative of v(x) is supposed to have v in the denominator, and I don't see how they got that answer...and then what? Where did the t go to get that final v max?

 

[tiny-tim]

Welcome to PF!

The roller coaster is designed according to safety regulations that prohibit the speed of the car from exceeding 20. Find the maximum value of A allowed by these regulations.
…
v = A * √[1 + 9t^2 * (t-4)^2]
…
I don't know how the do the last part of the equation at all. A is an unknown, but t is a variable as well...I tried using a hint in the program but it just tels me the answer should be vmax = A * sqrt [145]

I thought maybe, you find the derivative of v(x) and check when it's zero, then you know it's maximum velocity. But in the answer given, the derivative of v(x) is supposed to have v in the denominator, and I don't see how they got that answer...and then what? Where did the t go to get that final v max?

Hi theowne! Welcome to PF!

Hint: you're only asked for 0≤t≤4 …

the maximum may be at dv/dt = 0, or it may be at 0 or 4

EDIT: hmm … just noticed … you can see where the maximum is without any hard work …

v is a maximum when t(t - 4) is a maximum, isn't it?

 

[baba89]

I would first like to commend you on your attempt at solving the problem and finding the general expression for the speed of the car. Your approach seems to be correct and I agree with your answer for the first part of the question.

For the second part, we need to find the maximum value of A that satisfies the safety regulations, which state that the speed of the car should not exceed 20. To do this, we can set the expression for speed, v, equal to 20 and solve for A. This will give us the maximum value of A that satisfies the given condition.

So, let's set A * sqrt[1 + 9t^2 * (t-4)^2] = 20 and solve for A. This will give us A = 20/sqrt[1 + 9t^2 * (t-4)^2]. We can then plug in the value of t to get the maximum value of A allowed by the regulations. In this case, t = 4, so we get A = 20/sqrt[145]. This gives us the final answer of vmax = A * sqrt[145] = 20, as given in the hint.

I hope this helps you understand the problem and how to approach it. Keep up the good work!