Find Min Acceleration Experienced by Car Passengers on Elliptical Path

[the_dialogue]

A car moves around an ellipse with the equation
$$\frac{x^2}{60^2} + \frac{y^2}{40^2} = 1$$
-60<x<60
-40<y<40

The car keeps a constant speed of 60 km/h.

I have to find the minimum acceleration experienced by the passengers of the car.

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I am nearly sure that my solution contains the right process. However, they may be an error in calculation throughout. If anyone could check my answers, I would strongly appreciate it.

There is no tangential acceleration. Therefore, acceleration = normal acceleration. Normal acceleration = $$\frac{v^2}{p}$$
Where:
$$p = \frac{(1+ (dy/dx)^2)^\frac{3}{2}}{|d^2y/dx^2|}$$

After manipulation, I end up with a function for a:

$$a = 2400*(v^2)*(\frac{180^2-9x^2}{(180^2-5x^2)(60^2-x^2)})^\frac{3}{2}$$

Afterwards, I diffentiated this 'a' function and found its roots. x=0, x=42.43, and x=60. Then, i calculated the accelerations as needed.

Please tell me where I have gone astray!

 

[member 553083]

The minimum acceleration experienced by the passengers of the car is:a_min = 2400*(60^2)*(\frac{180^2}{(180^2-5*42.43^2)(60^2-42.43^2)})^\frac{3}{2} a_min = 0.827 m/s^2