Newton 2 as a stepping rule

Newton's second law for motion along a line ($x$) can be represented as a pair of differential equations:  the rate of change of the position is the velocity, and the rate of change of the velocity (the acceleration) is determined by the net force divided by the mass.

$$\frac{dx}{dt} = v\;\;\;\;\frac{dv}{dt}=\frac{F^{net}}{m}$$

Since we know that derivatives can help us predict the future by using them as stepping rules, let's see if we can use Newton 2 as a stepping rule to figure out how something is going to move.

In general, when we use a derivative as a stepping rule, we start on one side of the time interval and step to the end. Here's what it looks like for a general function of $x$, $f(x)$, when the derivative is given by a function $g(x)$ for starting with a value at $x_1$ of $g_1$ and a value $g_2$ to be found at a larger value of $x$, $x_2$.

$$ \frac{df}{dx} = g(x)$$

$$df = g(x) dx$$

$$f_2 - f_1 = g(x) (x_2 - x_1)$$

$$f_2 = f_1 + g(x) (x_2 - x_1)$$

For the two derivatives of Newton's second law we have to do this for each one.

$$ \frac{dx}{dt}= v(t) \;\;\;\;\;v(t) = \frac{F^{net}}{m}$$

$$dx = v(t) dt\;\;\;\;\;dv = (\frac{F^{net}}{m}) dt $$

$$x_2 - x_1 = v(t) (t_2 - t_1)\;\;\;\;\;v_2 - v_1 = \frac{F^{net}}{m} (t_2 - t_1)$$

$$x_2 = x_1 + v(t) (t_2 - t_1)\;\;\;\;\;v_2 = v_1 + \frac{F^{net}}{m} (t_2 - t_1)$$

The last set of equations shows us how to start at position $x_1$ and velocity $v_1$ at a time $t_1$ and find their values $x_2$ and $v_2$, at a slightly later time, $t_2$.

One small problem. The velocity in the left set of equations depends on time, and the forces in the right set of equations can depend on position and velocity, which depends on time. The analog to what we did for the general derivative is to evaluate everything on the left at the initial time. It turns out that while this is usually OK, for cases where our object is oscillating back and forth, the small errors in each step tend to add rather than cancel so the solution goes astray after a little while. Some fancier math shows that if one steps the velocity equation first rather than stepping the position equation, and then puts the new velocity into the position equation, things work better.  So here is our final stepping rule for solving a motion governed by Newton's second law along one dimension. Starting at time t₁ with position x₁ and velocity v₁, here's how to predict the position x₂ and the velocity v₂ at a later time, t₂ = t₁ + dt

$$v_2 = v_1 + \frac{F^{net}(x_1,t_1)}{m} (t_2 - t_1)$$

$$x_2 = x_1 + v_2 (t_2 - t_1)$$

Here's what it means conceptually. 

1. We start at some time ($t_1$) knowing the position and velocity ($x_1$ and $v_1$).
2. We figure out the net force at that instant (which might depend on the position or the velocity or both).
3. The acceleration air time $t_1$ is the net force divided by the mass of the object being considered. The change in velocity is the acceleration times the time interval.  This yields the velocity, $v_2$, at the later time, $t_2 = t_2 + dt$.
4. The change in position is the velocity times the time interval.  This yields the position, $x_2$ at the later time, $t_2$.
5. We can now take $t_2$ as our initial time and repeat the process, making a prediction for how the object will move in response to the forces it feels.

To see how to set this up on a spreadsheet and do this kind of calculation for yourself, see the follow-on page.

Joe Redish 9/17/11