0

I.e., what is the set of solutions to this system of differential equations:

$\frac{dx}{dt} = a_x t + v_{ox}$

$\frac{dy}{dt} = a_y t + v_{oy}$

which corresponds to the path of an object under constant acceleration in both the $x$ and $y$ directions, with initial velocities $v_{ox}$ and $v_{oy}$.

By integrating both then solving one equation to find $t = \pm \sqrt{\frac{a_x t^2}{2}}$ and plugging that into the other, I have found that when $v_{ox}=x_o=0$, the solution set is:

$y = \frac{a_y}{a_x}x \pm v_{oy}\sqrt{\frac{2 x}{a_x}} + y_o$

where $x_o$ and $y_o$ are the initial $x$ and $$ positions, respectively.

How can I solve this for the general case, i.e. when $v_{ox}$ and $x_o$ are not constrained to be zero?

Matthew
  • 227
  • In your system of equations, you need the derivatives of $v_x$ and $v_y$, not $x$ and $y$. – Raskolnikov Jul 23 '19 at 20:33
  • 1
    How do you know that $a_x\ne0$ when you solved for $t$ in the parametric equation for $x$ and then divided by it to get the solution in your question? If you want to eliminate $t$, then I think it would be better to use an implicit equation instead of trying to solve for $y$ in terms of $x$. – amd Jul 23 '19 at 22:06
  • 1
    The path will be parabolic. You might want to have a look at https://math.stackexchange.com/q/530310/265466 for ways to convert from a quadratic parameterization to the general conic equation of the parabola. – amd Jul 23 '19 at 22:14

2 Answers2

2

For an object with constant acceleration $\mathbf{a}$, initial velocity $\mathbf{v}_0$, and initial position $\mathbf{x}_0$, the trajectory of the particle can be found by integrating $d^2\mathbf{x}/dt^2 = \mathbf{a}$ twice: $$ \mathbf{x}(t) = \mathbf{a}\frac{t^2}{2} + \mathbf{v}_0 t + \mathbf{x}_0. $$ Now you could do some sort of awful quadratic formula thing to solve for $y$ in terms of $x$. But let's be more clever about this. First, take cross products with $\mathbf{a}$ and $\mathbf{v}_0$ to isolate the powers of $t$: $$ \mathbf{a}\times(\mathbf{x}-\mathbf{x}_0) = (\mathbf{a}\times\mathbf{v}_0)t \\ (\mathbf{x}-\mathbf{x}_0)\times\mathbf{v}_0 = (\mathbf{a}\times\mathbf{v}_0)\frac{t^2}{2}. $$ We then eliminate $t$ by taking the dot product of the first equation with itself and the second with $\mathbf{a}\times\mathbf{v}_0$, giving $$ |\mathbf{a}\times(\mathbf{x}-\mathbf{x}_0)|^2 = 2(\mathbf{a}\times\mathbf{v_0})\cdot[(\mathbf{x}-\mathbf{x}_0)\times\mathbf{v}_0], $$ which implicitly gives the trajectory. This is also coordinate-independent, so you don't need to worry about whether $\mathbf{a}$ happens to be parallel to one of the axes. You do still have to worry about the degenerate case $\mathbf{a}\times\mathbf{v}_0 = 0$, where the particle follows a ray rather than a parabola, but that's a pretty easy special case.

EDIT: To make it a little more clear that this is a parabola, we can use the expansion $(\mathbf{x}-\mathbf{x_0}) = [\hat{\mathbf{a}}\cdot(\mathbf{x}-\mathbf{x}_0)]\hat{\mathbf{a}} + \hat{\mathbf{a}}\times[(\mathbf{x}-\mathbf{x}_0)\times\hat{\mathbf{a}}]$, where $\hat{\mathbf{a}} = \mathbf{a}/a$, and do some vector algebra to get $$ \frac{a|\hat{\mathbf{a}}\times(\mathbf{x}-\mathbf{x}_0)|^2}{2|\hat{\mathbf{a}}\times\mathbf{v}_0|^2} +\frac{(\hat{\mathbf{a}}\cdot\mathbf{v}_0)(\hat{\mathbf{a}}\times\mathbf{v}_0)}{|\hat{\mathbf{a}}\times\mathbf{v}_0|^2}\cdot[\hat{\mathbf{a}}\times(\mathbf{x}-\mathbf{x}_0)]= \hat{\mathbf{a}}\cdot(\mathbf{x}-\mathbf{x}_0), $$ which is just the trajectory equation $$ y-y_0 = -\frac{g(x-x_0)^2}{2v_x^2} + \tan\theta(x-x_0) $$ written in coordinate-independent form.

eyeballfrog
  • 22,485
0

Correct would be writing Newton's equation in vector form:

$$\frac{d^2 \vec{r}}{dt^2}=\vec{a}$$

Then the general solution is:

$$\vec{r}=\vec{r}_0+\vec{v}_0 t+\vec{a}\frac{t^2}{2}$$

For $x$ and $y$ we have the same form of equations.

To find the trajectory, you only need to solve a quadratic equation for $t(x)$ and then plug it into the equation $y(t)$.

Can you solve this for $t$?

$$t^2+2 \frac{v_{0x}}{a_x} t+2 \frac{x_0-x}{a_x}=0$$

Yuriy S
  • 31,474