1D Action

@Karan Chatrath has shown how to derive the result. I will comment on the physical significance a bit more. When I form these kinds of problems, I use the following procedure:

1) Start with a standard physical trajectory, where a particle moves from $(x_1, t_1)$ to $(x_2, t_2)$
2) Deform the trajectory in such a way that the end points in space and time remain the same. This is usually done by adding some extra term with a "deformation parameter" ( $\alpha$ , in this case)

Since $\alpha$ can take any value, there are infinitely many conceivable paths with the same end points in space and time. But the vast majority of these are non-physical, and would never happen in reality. The physical paths correspond to:

$\frac{d S}{d \alpha} = 0$

Applying this to the action for this problem yields:

$\frac{d S}{d \alpha} = \alpha T = 0 \implies \alpha = 0$

This means the velocity has the form:

$v(t) = v_0$

In the absence of a potential (force), the physical path is the one that has constant velocity. So the principle of stationary action yields Newton's 1st law.

$S(\alpha) = \int_{0}^{T} \left(v_o + \alpha \sin\left(\frac{2 \pi t}{T}\right)\right)^2 \ dt$ The above is a straightforward evaluation which leads to the answer:

$S(\alpha) = v_o^2T + \alpha^2\frac{T}{2}$

I find this situation paradoxical. On one hand, it is easy to relate this to Newton's first law as no external force acts on the system. On the other hand, the acceleration of the particle is:

$a(t) = \frac{2 \pi \alpha}{T} \cos\left(\frac{2 \pi t}{T}\right)$

So, here is a case where there is no external force but a non-zero acceleration. What causes the particle to accelerate?

$\frac{1}{2}mv^2=\frac{1}{2}m*v(t)^2$

Given that $v(t)=v_0+\alpha sin(\frac{2\pi t}{T})$ , we can plug this into the right side of the above equation to give:

$\frac{1}{2}mv^2=\frac{1}{2}m (v_0^2+\alpha ^2 sin^2(\frac{2\pi t}{T}))$

To find the "action," which I assume means the kinetic energy, you need to integrate $\frac{1}{2}mv^2$ , meaning we need to integrate both sides.

Note: The upper limit of the left-hand side(LHS) is $v_0$ , and the upper limit of the right-hand side(RHS) is $T$ .

$\int_{0}^{v_0} \frac{1}{2}mv^2 \,dt = \int_{0}^{T} \frac{1}{2}m (v_0^2+\alpha ^2 sin^2(\frac{2\pi t}{T})) \,dt$

The LHS comes out to $m(v-v_0)$ . The RHS comes out to:

$\frac{mT(2v_0^2+\alpha ^2)}{4} - \frac{\alpha ^2 m T sin(4 \pi)}{16 \pi}$

Simplifying this^ gives:

$\frac{mT}{4} \left( (2v_0^2 + \alpha ^2) - \frac{\alpha ^2 sin(4 \pi)}{4 \pi} \right)$

We know that $sin(4 \pi)=0$ , so the second term in the above equation will be zero, leaving this:

$\frac{mT}{4} (2v_0^2 + \alpha ^2)$

We also know that $m=2$ , so plugging that value into the above equation gives:

$v_0^2 T + \alpha ^2T/2$

So now that we have this, we can see that $a=2, b=2, c=2$ , meaning $a+b+c=6$ .