Finding Coordinates on a circle with time/speed

[bsmithysmith]

Marla is running clockwise around a circular track. She runs at a constant speed of 3 meters per second. She takes 46 seconds to complete one lap of the track. From her starting point, it takes her 12 seconds to reach the northernmost point of the track. Impose a coordinate system with the center of the track at the origin, and the northernmost point on the positive y-axis. [UW]
a) Give Marla’s coordinates at her starting point.
b) Give Marla’s coordinates when she has been running for 10 seconds.
c) Give Marla’s coordinates when she has been running for 901.3 seconds.

So from what I understand, it's 3 meters/second and it takes 46 seconds to complete a lap, so \(\displaystyle 46*3 = 138 meters\) and then it's a circle, so 1 revolution is \(\displaystyle 138/2Pi\), giving the radius as approximately 21.96. From there, I have no idea what to do. I'm thinking angular speed \(\displaystyle (w = distance/time)\) and linear speed \(\displaystyle (v = arclength/time)\) although I think the linear speed is given as 3 meters/second, so angular speed is \(\displaystyle w = (21.96*theta)/seconds\). The teacher did this in class but sadly his back was towards me the entire time so I got nothing from class that day. (Crying)

 

[MarkFL]

Like you, the first thing I would do is find the radius of the track:

\(\displaystyle 3\cdot46=2\pi r\implies r=\frac{69}{\pi}\)

I would then consider a parametric description of her position on the track. A parametric description of a circle of radius $r$ is:

\(\displaystyle x=r\cos\left(\theta_0+\omega t \right)\)

\(\displaystyle y=r\sin\left(\theta_0+\omega t \right)\)

Since Marla is running in a clockwise direction, we want to negate the parameter $t$. Also, we are told the period in seconds is 46, hence:

\(\displaystyle \omega=\frac{2\pi}{46}=\frac{\pi}{23}\)

So now we have:

\(\displaystyle x=r\cos\left(\theta_0-\frac{\pi}{23}t \right)\)

\(\displaystyle y=r\sin\left(\theta_0-\frac{\pi}{23}t \right)\)

Now, we need to determine the initial angle $\theta_0$. We are tole she is 12 seconds from:

\(\displaystyle \theta=\frac{\pi}{2}\)

\(\displaystyle \theta_0-\frac{12\pi}{23}=\frac{\pi}{2}\)

\(\displaystyle \theta_0=\frac{\pi}{2}+\frac{12\pi}{23}=\frac{47}{46}\pi\)

And so we have:

\(\displaystyle x=r\cos\left(\frac{47}{46}\pi-\frac{\pi}{23}t \right)\)

\(\displaystyle y=r\sin\left(\frac{47}{46}\pi-\frac{\pi}{23}t \right)\)

Using the value we found for $r$ and rewriting the arguments for the trig. functions by factoring, we may state:

\(\displaystyle x=\frac{69}{\pi}\cos\left(\frac{\pi}{46}\left(47-2t \right) \right)\)

\(\displaystyle y=\frac{69}{\pi}\sin\left(\frac{\pi}{46}\left(47-2t \right) \right)\)

 

[bsmithysmith]

\(\displaystyle x=r\cos\left(\theta_0+\omega t \right)\)

\(\displaystyle y=r\sin\left(\theta_0+\omega t \right)\)

This part was confusing. The theta looks like it has a zero subscript and the omega, which is theta/time, is multiplied by time again? Sorry to ask a lot of questions; I'd really like to understand the problem more.

 

[MarkFL]

$\theta_0$ represents the initial angle (when $t=0$) and $\omega$ represents the angular speed in radians per second. It determines the period of the parametric functions. If $T$ is the given period, then we know:

\(\displaystyle \omega=\frac{2\pi}{T}\)

 

[CellCoree]

a) Marla's starting point is at the origin, so her coordinates are (0,0).
b) After 10 seconds, Marla has completed 10/46 or approximately 0.2174 laps around the track. This is equivalent to approximately 30.15 degrees of rotation. Using the formula for polar coordinates (r,θ), her coordinates would be (21.96, 30.15).
c) After 901.3 seconds, Marla has completed 901.3/46 or approximately 19.587 laps around the track. This is equivalent to approximately 2717.4 degrees of rotation. Using the formula for polar coordinates (r,θ), her coordinates would be (21.96, 2717.4). However, this point is not on the track as Marla would have completed multiple laps. To find the coordinates on the track, we need to find the equivalent angle within one lap. This can be done by taking the remainder of 2717.4 divided by 360 (the number of degrees in one lap). This gives us an equivalent angle of approximately 77.4 degrees. So Marla's coordinates at this time would be (21.96, 77.4).