When is $\vec{r}(t)=\vec{r}_0+\vec{v}_0t+\frac{1}{2}\vec{a}_0t^2$ an equation of a straight line in 3D space?

Question

Under what conditions $$\vec{r}(t)=\vec{r}_0+\vec{v}_0t+\frac{1}{2}\vec{a}_0t^2$$ is an equation of a straight line in 3D space if $\vec{r}_0,\vec{v}_0$ and $\vec{a}_0$ are all constant vectors?

In the special case when $\vec{a}_0=0$, I get a straight line eq. by eliminating $t$: $$\frac{x-x_0}{v_{0x}}=\frac{y-y_0}{v_{0y}}=\frac{z-z_0}{v_{0z}}.$$

Answer 1

If $\vec{v}_0$ and $\vec{a}_0$ are both $\vec{0}$, we get a point. If $\vec{v}_0\ne0$ and $\vec{a}_0=\lambda\vec{v}_0$, the locus is $\vec{r}_0+(t+\frac12\lambda t^2)\vec{v}_0$. If $\vec{v}_0\ne0$ and $\vec{a}_0$ isn't of the above form, the path isn't a straight line because its time derivative $\vec{v}_0+\vec{a}_0t$ has $t$-dependent direction.

Answer 2

What is being described is the motion of some object with an initial position $\vec r_0$, initial velocity $\vec v_0$, and a constant acceleration $\vec a_0$. If you think about this for a while, it may become apparent that the condition for this to be a straight line is that $\vec a_0$ is parallel to $\vec v_0$. Here is a short explanation of why:

What you want is for the displacement from $\vec r_0$ to be a straight line. This displacement is given by $\vec s(t) = \vec r(t) - \vec r_0 = \vec v_0 t + \tfrac 12 \vec a_0 t^2$.

Since $\vec s(0) = \vec 0$, the condition for $\vec s$ to be a straight line is that $\vec s$ should be a scalar multiple of some vector (say $\vec n$) for all $t$.

Now if $\vec s(1) = \vec v_0 + \tfrac 12 \vec a_0 = k_1 \vec n$ is a scalar multiple of $\vec s(2) = 2\vec v_0 + 2 \vec a_0 = k_2 \vec n$, then so are $\vec s(2) - 2\vec s(1) = (k_2 - 2k_1)\vec n = \vec a_0$ and $2\vec s(1) - \tfrac 12 \vec s(2) = (2k_1 - \tfrac 12 k_2)\vec n = \vec v_0$.

So $\vec a_0$ must be parallel to $\vec v_0$. This is certainly sufficient as we then get $\vec s(t) = (2k_1 - \tfrac 12 k_2)\vec nt + \tfrac 12(k_2 - 2k_1)\vec n t^2$ is some scalar multiple of $\vec n$.

You could also justify this with calculus or various bits of knowledge of vectors and matrices, but I've tried to present a simple argument.

Note that strictly we get $\vec s(t) = (2k_1 t - \frac 12 k_2 t + \frac 12 k_2 t^2 - k_1 t^2) \vec n$. This will be identically zero if $k_1 = k_2 = 0$, which happens if $\vec v_0 = \vec a_0 = \vec 0$. So we need this to not be the case in order for the shape not to be a point.

Answer 3

The line equation says

$$ p(t) = p_0+t\vec v $$

now arranging

$$ r(t) = r_0 + t(\vec v_0+\frac t2\vec a_0) $$

if

$$ \{\vec v_0 = 0\cup\vec a_0 = 0\cup\vec a_0 =\lambda \vec v_0 \} $$

but not simultaneously, we have that $r(t)$ describes a line. Here $\lambda\ne 0$

Answer 4

WLOG set $\vec r_0=0$;

Assume $\vec v_0$, $\vec a$ both $\not =0$, and not parallel.

Then $\vec v_0$ and $\vec a$ span a plane.

The problem is now 2D, $\vec v$ and $\vec a$ are 2 vectors in the $xy$-plane.

$\vec d(t):= t\vec v_0 +(1/2)t^2 \vec a$ is the diagonal vector of the parallelogram with base $t\vec v_0$ and side vector $(1/2)t^2 \vec a$.

$\vec d(t)$ changes in direction as a function of $t$, since the components change at a different rate.

Hence not a straight line.