Track and Field Car

Simple basic geometry this time.

A Ferrari Testarossa is 4485 mm long and 1130 mm wide. It's taken to a track and field area where the field inside the track consists of a square in the middle and two semicircles at the ends, with a total perimeter length of 400 meters. How wide does the track need to be for the car to be able to go a full lap?

Are we to assume there is a wall on the inside and outside perimeter of the track, that the car is a perfect rectangle, and that the car's turning radius is irrelevant?

Definitions:

r = radius of semi-circle at end of track
L = 4485mm (length of car)
w = 1130mm (width of car)
w' = width of track
r' = r + w'


Calculation:

r = 400m / (2pi + 4) ~= 38898.5mm
theta = tan-1( (L / 2) / (r + w) )
r' = (L / 2) / sin(theta)
w' = r' - r

Plugging all of that into wolfram alpha (because I'm lazy)

w' ~= 1192.77mm

*edit*
For this to actually work, the car would need to be able to steer both the front and rear wheels

nice work, and yes, the 400 meters is as you wrote it, and yes, I thought that the Ferrari looks like a moving rectangle when seen from above, either with four-wheel steering or with a very slippery track and no steering needed at all.

I didn't use trigs.

Distance from the center of the semicircle to the middle of left side of the car (if driving counter-clockwise) = R.

Distance from the center of the semicircle to the right front / back corner = ((R+w)^2+(L/2))^(1/2)

So the width is of the track is ((R+w)^2+(L/2))^(1/2) - R.

which gives exactly the same result, of course.