MHB 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 $$46*3 = 138 meters$$ and then it's a circle, so 1 revolution is $$138/2Pi$$, giving the radius as approximately 21.96. From there, I have no idea what to do. I'm thinking angular speed $$(w = distance/time)$$ and linear speed $$(v = arclength/time)$$ although I think the linear speed is given as 3 meters/second, so angular speed is $$w = (21.96*theta)/seconds$$. 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:

$$3\cdot46=2\pi r\implies r=\frac{69}{\pi}$$

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

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

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

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:

$$\omega=\frac{2\pi}{46}=\frac{\pi}{23}$$

So now we have:

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

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

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

$$\theta=\frac{\pi}{2}$$

$$\theta_0-\frac{12\pi}{23}=\frac{\pi}{2}$$

$$\theta_0=\frac{\pi}{2}+\frac{12\pi}{23}=\frac{47}{46}\pi$$

And so we have:

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

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

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

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

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

 

[bsmithysmith]

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

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

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:

$$\omega=\frac{2\pi}{T}$$