The quadratic variation of kinetic energy with velocity can be explained by the symmetry properties of space and time. The lagrangian function is defined as $\mathcal{L}=T-U$, where $T$ is the kinetic energy and $U$ is the potential energy.

We know that space is homogeneous and isotropic, and time is homogeneous. For a free particle, it follows that the lagrangian $\mathcal{L}$ should have the following properties:

1. $\mathcal{L}$ should not depend on the position coordinate.
2. $\mathcal{L}$ should not depend on the velocity vector. Rather it should depend on the magnitude of the velocity, i.e., some power of the velocity vector.
3. $\mathcal{L}$ should not depend on the time coordinate.

So the general form of the lagrangian for a free particle is $$\mathcal{L}(x,v,t)=\alpha v^n$$ where $\alpha$ is a constant independent of the coordinates, velocities and time. Now, the momentum can be calculated by using the relation $$p=\frac{\partial\mathcal{L}}{\partial v}=\alpha nv^{n-1}$$ However, momentum is always a linear function of velocity (which can be proved by dimensional analysis). This is possible only when $n=2$ in the above expression.

Since we are considering a free particle (which has only kinetic energy), the lagrangian (choosing $n=2$) is $$\mathcal{L}=T=\alpha v^2$$ Thus, the kinetic energy is proportional to $v^2$ and not any other power of $v$.

The quadratic variation of kinetic energy with velocity can be explained by the symmetry properties of space and time. The lagrangian function is defined as $\mathcal{L}=T-U$, where $T$ is the kinetic energy and $U$ is the potential energy.

We know that space is homogeneous and isotropic, and time is homogeneous. For a free particle, it follows that the lagrangian $\mathcal{L}$ should have the following properties:

1. $\mathcal{L}$ should not depend on the position coordinate.
2. $\mathcal{L}$ should not depend on the velocity vector. Rather it should depend on the magnitude of the velocity, i.e., some power of the velocity vector.
3. $\mathcal{L}$ should not depend on the time coordinate.

So the general form of the lagrangian for a free particle is $$\mathcal{L}(x,v,t)=\alpha v^n$$ where $\alpha$ is a constant independent of the coordinates, velocities and time. Now, the momentum can be calculated by using the relation $$p=\frac{\partial\mathcal{L}}{\partial v}=\alpha nv^{n-1}$$ However, momentum is always a linear function of velocity which can easily be proved by dimensional analysis. This is possible only when $n=2$ in the above expression.

Since we are considering a free particle (which has only kinetic energy), the lagrangian (choosing $n=2$) is $$\mathcal{L}=T=\alpha v^2$$ Thus, the kinetic energy is proportional to $v^2$ and not any other power of $v$.