Position with respect to time with friction

Question

I'm interested in equations of motion when friction is present for a little graphical side project I am working on. I'm really rusty with physics, so I apologize if this is a basic question or doesn't make sense...

Consider an object on a plane in three dimensions. It has initial position $ \vec{x_0} $ on the plane and initial velocity $ \vec{v_0} $ tangential to the plane. Assume it has acceleration (i.e. downhill gravity), $ \vec{a_G} $, which is constant and tangential to the plane. Now assume it has an additional acceleration (from friction, which I'm modeling as an acceleration), $\vec{a_F}(t) $, with constant magnitude $ k $ but variable direction, opposite that of velocity.

Is there a function $ \vec{x}(t) $ that gives position with respect to time? Something in the vein of the $$ x(t) = x_0 + v_0*t + 0.5*a*t^2 $$ that I learned in high school?

Answer 1

In one dimension (it is easier that way) we have $$x''=a-kx'\\kx'+x''=a\\x'=\frac ak+ce^{-kt}\\c=v_0-\frac ak\\x=\frac akt-\frac {1}k\left(v_0-\frac ak\right)e^{-kt}+x_0$$

Added: the above misreads the problem to involve a frictional force proportional to the velocity. If the friction force is constant in magnitude, then as long as the velocity is in the same direction as the external acceleration we will just have $$x=x_0+v_0t+\frac 12(a-f)t^2$$

Answer 2

Constant force in the direction opposite to velocity complicates the problem.

Let's define $x$ as the direction directly down the place and $y$ as the horizontally within the plane.

Then:

$$a_{x} = a_{G} -k \hat{v}_{x} $$ $$a_{y} = -k \hat{v}_{y} $$

Now:

$$ \hat{v}_{x} = \frac{v_{x}}{|v_{x}^2+v_{y}^2|} $$

with likewise for $\hat{v}_{y}$.

As you can see, this couples the $x$ and $y$ accelerations. I'd suggest you go with the plan of working with the accelerations directly and just doing step-wise calculations.