I have a car traveling on an elliptical race track at a constant angular velocity of A radians/sec. The angular velocity is calculated at the intersection of semi-major & semi-minor axis. I know the eccentricity e, semi-major axis a, semi-minor axis b of the ellipse & the time T it takes for completing one lap of the race track.
I am not much familiar with calculus related to elliptical shapes. For a circle, the distance traveled can be found through Speed-Distance-Time formula. But for an ellipse, to maintain constant angular velocity, the linear velocity needs to be changed continuously. The car needs to go slower around semi-minor axis point & faster around semi-major axis point as far as I know.
I tried to find how the linear velocity should change but couldn't make sense of it. I also tried to approximate the distance measurement using the general formula
$$ \frac{Lap \ Circumference}{Lap \ Period} = \frac{Distance \ Travelled }{Time \ to \ travel\ Distance \ length}$$
I am looking for a formula to calculate this.
How to find the actual distance in meters traveled by the car after time t. Assume that the start/finish line is at semi-major axis point.
HINTS:
Since $ \theta = \omega t $ where $\omega$ is constant we can do the integration w.r.t. $\theta$ or time. In central ellipse polar form
$$ \frac{\cos^2 \theta }{a^2}+\frac{\cos^2 \theta }{b^2}= \frac{1}{r^2} \tag1 $$
Differentiate w.r.t. arc ( primed w.r.t. arc $s$ )
$$ {\theta^{'}} \,\sin 2 \theta (1/b^2-1/a^2) =\frac{-2{r^{'}} }{r^3} \tag2 $$
Distance along arc can be further found by evaluating through radius $r$
$$ ds^2=dr^2 + (r\, d \theta)^2; \quad \tag 3 $$
$${r{'} ^2 + {({ r\theta^{'}} )^{2}} } = 1 \tag4 $$
Solve for $ ( r^{'}, \theta^{'}) $ from (2) and (3) to form two coupled differential equations and solve for $r, \theta= \omega t. $