Question

A rocket car is constrained to move on an elliptical track (semi major axis p and semi minor axis q). The car is moving at a constant speed mu. Determine the acceleration of the car is meters per second squared. P = 4 km, q = 2km and mu = 360 km/hr.

Answer 1

Finding the value of R is a calculus challenge. Our ellipse is defined as such:

x^2/a^2 + y^2/b^2 = 1

where a and b are the half-width and half-length of the ellipse.

Re-arrange for y:
y = b*sqrt(1 - x^2/a^2)

Recall the calculus origin of curvature:
kappa = (d^2 y/dx^2)/(1 + (dy/dx)^2)^(3/2)----(1)

Take the first derivative of the ellipse:

dy/dx = -b*x/(a^2*sqrt(1-x^2/a^2))

Take the second derivative:
d^2 y/dx^2 = -b*x^2/(a^4(1-x^2/a^2)^(3/2)) - b/(a^2*sqrt(1-x^2/a^2))
substitute in (1)

By definition of curvature
| kappa | = 1/R

R=1/kappa
Now, use the centripetal acceleration formula:
a = v^2/R

Data:
v:=360km/h;

Result is for centripetal acceleration, which is also the total acceleration because speed is constant:
Note: as x and y are not given, expression for R would be very complicated, and a would be further complicated. Hence, I have omitted these out. It is simple substitution in subsequent equations. If x and y are given, solution can be simplified easily.