Car's Trajectory on a Roller Coaster Track

[ProBasket]

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 $$A$$ is a positive dimensionless constant.

I have two questions that i don't get:

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

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 $$A$$ 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

 

[MathStudent]

in general,, if $\vec{r}$ is a vector equation s.t.

$$\vec{r} (t) = f(t) \vec{x} + g(t) \vec{y} + h(t) \vec{z}$$

then the derivative of $\vec{r} (t)$ wrt t is given by,

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

where $\vec{x}, \ \vec{y}, \ \vec{z}$ are the unit vectors along the x, y, z, axis respectively.

if $\vec{r}(t)$ is a vector vlalued function for the position of a particle, then $\frac{d}{dt} \vec{r} (t)$ 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 )

 

[wais]

1) Yes, you are correct in that the derivative of the position function with respect to time will give you the velocity function. So, the general expression for the speed v of the car would be: v(t) = 6at - 12, in meters per second.

2) To find the maximum value of A allowed by the safety regulations, we need to set the maximum speed of the car (20m/s) equal to the speed function we derived in part 1. So, we have: 20 = 6at - 12. Solving for A, we get A = (20 + 12)/6t = 32/6t. Since t is a positive value (time cannot be negative), the maximum value of A allowed by the regulations would be when t is at its minimum value of 0. So, A = 32/6(0) = 0. Therefore, A should be less than or equal to 0 for the car to stay within the safety regulations.