A particle of mass $m = 2$ travels along a one-dimensional path from $x = 0$ to $x = L$ with the following velocity:

$v(t) = v_0 + \alpha \sin(2 \pi t / T) \\ T = \frac{L}{v_0}$

Regardless of the particular value of $\alpha$ , the particle starts at $(x_1, t_1) = (0,0)$ and ends at $(x_2, t_2) = (L,T)$ . Since there is no potential, the action for this path as a function of $\alpha$ is:

$S(\alpha) = \int_0^T \frac{1}{2} m \, v^2(t) \, dt = v_0^a \, T + \alpha^b \, \frac{T}{c}$

In the above expression, $a,b,c$ are positive integers. What is $a + b + c$ ?

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Bonus:
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Which one of Newton's Laws does this result relate to?

The answer is 6.

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@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.