I encountered a Physics Olympiad problem:
A ball bearing rests on a ramp fixed to the top of a car which is accelerating horizontally. The position of the ball bearing relative to the ramp is used as a measure of the acceleration of the car. Show that if the acceleration is to be proportional to the horizontal distance moved by the ball (measured relative to the ramp), then the ramp must be curved upwards in the shape of a parabola.
Attempt at a solution:
Let $A$=magnitude of acceleration of the car w.r.t a stationary observer, $a$=magnitude of acceleration of of the ball bearing w.r.t the stationary observer and $a'$=magnitude of acceleration of the ball bearing as observed in the accelerating car frame, then $$A=\mathcal{k}x' \tag{1}$$
where $x'$ is the horizontal position of the ball bearing as measured in the accelerating frame, with $x'=0$ at the bottom end of the ramp.
Since the ramp is a parabola, the position function of the ball bearing in the accelerating frame, should take the form $$y'=\alpha (x')^2 \tag{2}$$ where $\alpha$ is some constant and $y'$ is the vertical position as measured in the accelerating frame.
Using the definition of the fictitious force/acceleration $$\mathbf{a'=a-A} \tag{3}$$ The LHS of $(1)$ reads $$a+a'=kx'\tag{4}$$ (positive sense towards the direction of $\mathbf{A}$). But this doesn't seem to provide any useful information since I cannot define $a'$.
Should I try something like $\frac{dy'}{dx'}=\frac{\dot{y}}{\dot{x}}$? I am lost because I do not know whether I should analyse the ball bearing in its equilibrium (i.e. $A=constant, x'=constant$) to obtain some expressions for $y$ or it as a function $A(t)=kx'(t)$?
The diagram for the forces from elementary physics is useful here. (Car's acceleration towards the right).
For the vertical axis, the reaction of the surface, of module $N$ (no friction, no movement wrt the car), and gravity force, of module $mg$, must obey to $N\cos\alpha=mg$, with $\alpha$ the inclination of the ramp at the point where the balance occurs. For the horizontal axis $N\sin\alpha=ma$ because the ball is accelerating at the same amount than the car. So is,
$$a=g\tan\alpha$$
But $\tan\alpha=f'$, because the tangent of the inclination is the derivative of the function describing the profile of the ramp. So, $a=gf'$, the acceleration is proportional to the derivative.
We need that the distance from some point has to be proportional to the acceleration: $x=ka$, then the derivative must be proportional to $x$
$$Kx=f'$$ for some $K$
$$f(x)=Kx^2/2+C$$
A parabola