Motion in a plane situation There is something weird I find about the following situation. Suppose a particle has the $X$-coordinate $= 2+2t+4t²$ and $Y$-coordinate $= 4t+8t²$. So it's velocity in $X$ is $2+8t$ and velocity in $Y$ is $4+16t$. Acceleration in $X$ is $8$ and acceleration in $Y$ is $16$. Now I know this will follow a parabolic path in the $XY$ plane, but there are some relations we can obtain, namely $\frac{V_y}{V_x}=2$ and similarly $\frac{A_y}{A_x}=2$.
Does this mean that the velocity and acceleration act along same line? What's the significance of the ratio of velocities being independent of time?
I feel like there ought to be something special about this motion but I just can't point my finger at it.
 A: Summary
In this situation the particle is constrained to move in a straight line. You can confirm this by plotting the trajectory.
Details
Consider the velocity vector $\vec{v}(t) = \pmatrix{V_x \\ V_y}$ subject to the constraint $2 V_x - V_y = 0$. Now imagine the particle is in some sort of track and there is a constraint force $\vec{F} = \pmatrix{F_x \\ F_y}$ keeping it on the track. This force does no work and so the power of the force must be zero
$$ P = \vec{F} \cdot \vec{v} = F_x V_x + F_y V_y =0 $$
So if the force is proportional to $ \vec{F} = F \pmatrix{2 \\ -1}$ it would produce no power and it would act like the constraint force of the track.
$$ P = \vec{F} \cdot \vec{v} = 2 F V_x - F V_y = F (2 V_x - V_y) = 0 $$
So the particle moves along a track described by the normal (perpendicular) vector $$\vec{n} = \pmatrix{ 2  \\ -1 }$$
But since the normal vector is constant the track must be a straight line (since any curve would bend the normal vector).
A particle constrained on a straight line will always have velocity and acceleration vectors tangent to the line.
You can decompose the motion along the track, using the tangent vector $\vec{e} = \pmatrix{ 1 \\ 2 }$ and stating
$$ \vec{v}(t) = \vec{e}\, v(t) $$
$$ \require{cancel} \vec{a}(t) = \frac{\vec{e}}{\| \vec{e} \|}\, \dot{v}(t) + \frac{\vec{n}}{\| \vec{n} \|} \cancel{ \frac{v^2}{\rho }} $$
The reason the second part cancels is because the radius of curvature $\rho = \infty$.
A: When you say $\frac{v_y }{v_x} = 2$ or $\frac{a_y }{a_x} = 2$ you automatically refer to the ratio of the magnitudes of velocity and acceleration. This is because division of vectors is not defined in kinematics.So from these relations you cannot make any conclusion about the direction of acceleration or velocity.
A: One thing worth noticing is that we can prove the motion is linear pretty directly. We have two equations:
$$X = 2+2t+4t^2$$
and
$$Y = 4t+8t^2$$
If we divide the $Y$ equation by 2, we get
$$Y/2 = 2t+4t^2$$
Substituting into the $X$ equation, we get
$$X = 2 + Y/2$$
or
$$Y = 2X - 4$$
which is the equation of a line.
Another thing to notice is that an acceleration can only change the direction of velocity if it is not parallel with that velocity. There must be a component of the acceleration that points to the side of the velocity to change its direction. When you find that $V_X/V_Y = A_X/A_Y = 2$, you are proving exactly that the acceleration and velocity are parallel.
