# Physics with Calculus/Mechanics/Motion in Two Dimensions
## Motion with Constant Acceleration

In some situations, you will find that a particle is accelerating at a constant rate, so it is helpful to derive some basic relationships between the parameters of the problem. The problem of a constantly accelerating particle is completely characterized by the total time traveled, the initial position, the initial velocity, and acceleration. That is, we can figure out anything about particle at any time with just these quantities.
We will start from the definition of the problem that the particle has constant acceleration:
${\frac {d^{2}r}{dt^{2}}}=a=const.$.
Integrating once, we get
${\frac {dr}{dt}}=at+C$.
By the definition of velocity, we have
$v=at+v_{0}$
(I have replaced C with $v_{0}$ the initial velocity because at t = 0, $C=v(0)$ which is a constant, and it makes sense to call it $v_{0}$).
Integrating again, we have
$r={\frac {1}{2}}at^{2}+v_{0}t+r_{0}$
where $r_{0}$ is the initial position.
Now we can find position and velocity as functions of time, which is the complete solution to the problem. However, it is often useful to introduce the distance traveled as an independent variable instead of time. To do this, all we have to do is eliminate t from two equations above. The result is
$v^{2}=v_{0}^{2}+2a(r-r_{0})$.
Much less useful, but standardly taught is to use the average velocity as an independent variable. The average of a function between a and b is defined as
${\frac {1}{b-a}}\int _{a}^{b}f(x)dx$.
This is the limit of the arithmetic mean as the number of arguments goes to infinity.
Average velocity (averaged over time) is thus
${\bar {v}}={\frac {r-r_{0}}{t}}$.
eliminating r and $r_{0}$ with the equations of a and v as functions of time we get
${\bar {v}}={\frac {1}{2}}(v+v_{0})$.
So, here's the list of results:
$v=at+v_{0}$
$r={\frac {1}{2}}at^{2}+v_{0}t+r_{0}$
$v^{2}=v_{0}^{2}+2a(r-r_{0})$
${\bar {v}}={\frac {r-r_{0}}{t}}$
${\bar {v}}={\frac {1}{2}}(v+v_{0}).$
I could come up with more independent variables (it's very easy!), but generally it's easier just to do the calculation when you need it, and not to memorize tables and tables of simple results. On a similar note, you could calculate analogous results for a constant third derivative of position (commonly called jerk) or even constant nth derivative. Or if the nth derivative equals a given function, such as acceleration $=ksin(cosh(t/t_{0}))$. It all reduces to doing a couple messy integrals, then some messy algebra. The point is that the only reason working out constant acceleration is worth doing is because it comes up a lot, otherwise it is worthless.

## Motion in More Than One Dimension

It is an intriguing result that the motion of a particle in one direction does not affect the motion in any perpendicular direction. The classic example is if you shoot a gun level to the ground and drop a bullet at the same time, they hit the ground at the same time as if they started at the same height. That is, the motion of the bullet horizontally has absolutely no affect on how it moves vertically. You might ask why is it then that a paper airplane thrown and dropped do not hit the ground at the same time -- the answer is that the situation is fundamentally different because you have interaction with the air.
A mathematically precise way of saying this is that the velocity really is a vector. It adds like a vector, and you can split it up into components like a vector.
Using this notion, let's derive the constant acceleration equations for vectors and you can see how vectors are very useful.
${\frac {d^{2}\mathbf {r} }{dt^{2}}}=\mathbf {a} .$
That is,
${\frac {d^{2}r_{i}}{dt^{2}}}=a_{i}$
where the subscripts indicate the ith component of the vector.
Integrating each component (or equivalently, integrating the vectors),
$\mathbf {v} =\mathbf {a} t+\mathbf {v_{0}}$.
It's easy to see that everything is working out in exactly the same way it did for the 1 dimensional case, except we're doing it in every component!
Thus we have
$\mathbf {r} ={\frac {1}{2}}\mathbf {a} t^{2}+\mathbf {v_{0}} t+\mathbf {r_{0}}$
${\bar {\mathbf {v} }}={\frac {1}{2}}(\mathbf {v} +\mathbf {v_{0}} )$
${\bar {\mathbf {v} }}={\frac {\mathbf {r} -\mathbf {r_{0}} }{t}}$
and with some messy vector algebra,
$v^{2}{\hat {\mathbf {a} }}=v_{0}^{2}(2{\hat {\mathbf {v_{0}} }}\cdot {\hat {\mathbf {a} }}{\hat {\mathbf {v_{0}} }}-{\hat {\mathbf {a} }})+2\|\mathbf {a} \|(\mathbf {r} -\mathbf {r_{0}} )+2\|\mathbf {v} \|\|\mathbf {v_{0}} \|({\hat {\mathbf {v} }}\cdot {\hat {\mathbf {v_{0}} }}{\hat {\mathbf {a} }}-{\hat {\mathbf {v} }}\cdot {\hat {\mathbf {a} }}{\hat {\mathbf {v_{0}} }})$.
which simplifies down to:
$\mathbf {v} \cdot \mathbf {v} =\mathbf {v_{0}} \cdot \mathbf {v_{0}} +2\mathbf {a} \cdot (\mathbf {r} -\mathbf {r_{0}} )$.
But that's what it is if you were wondering. A hat above a vector is the vector divided by its magnitude, making it a unit vector in the direction of the original vector.