# Derivation of Kinematic Equations

To begin the derivation of the kinematic equations, first start with constant velocity (tumble buggie lab).  The more of this you can get the students to do, the better.

Figure 1 shows a plot of position vs time:

With the graph, guide the students to explain what the slope of the line represents.  You need to get the students to say the slope represents how much the position of the object changes every second, not just speed/velocity (we’re trying to build a concept/model, not memorize a word).  Continue to dialogue by asking questions such as “what does it mean if the line is higher/lower/negative,” and “what does it mean if the y-intercept is higher/lower” will help clarify the concept of velocity.

You can then go on to have the student explain how to determine the average velocity by adding an initial and final time (and corresponding position at said times) as seen in Figure 2.

Have the students start from the the beginning again:

$\large \overline{v}=\frac{\Delta x}{\Delta t}$
$\large \overline {v} = \frac {x_f – x_i}{\Delta t}$

Rewriting in $y=mx+b$ format, you get:

$x_f = \overline{v}\Delta t + x_i$

Also remind the students about the velocity vs time graph for constant motion, as seen in Figure 3.

From there, you can guide the students to the idea that the area under the line represents the displacement, shown in Figure 4:

Get students to explain that the shaded area is a rectangle:

$Area = base*height$

Since the “base” has units of time and the “height” has units of velocity, then:
$\large s*\frac{m}{s}=m$

So, the area has units of distance.  Since that area could be above or below the x-axis (thus positive or negative), the area is the displacement:

$\Delta x=\Delta t * v_o$

From this point, we can now engage the students with the velocity vs time graph for the cart on the inclined track (Figure 5).  Again, lead the students to say that this graph now shows that the velocity is changing with time.  Repeat the same questions about what the slope represents, what does it mean if line is steaper, etc.
{It may seem redundant, but a major misconception is what slope actually is. Most students only think of the definition/equation, not the true, physical concept}

Once again, lead the students to the definition of acceleration {I’ll omit the derivation to save time, I’m surprised you’re still reading this far}

$\overline {a} = \frac {\Delta v}{\Delta t}$
Which, again, can be rearranged into a “y = mx + b” format:

$v_f = v_i + \overline{a}\Delta t$

Which I’ll call Equation 1
{At this point, I’ll let you know that I circle the equation with a red marker in my class.  The only time I use a red marker is for a fundamental equation.  I picked this up from a college professor.  This fits in with the ABC Gum Rule talked about in Day 2.}

From here, we again lead the students to explain what the area under the velocity vs time graph represents.  The catch here is for them to recognize that the area under the curve can be broken into two geometric shapes as seen in Figure 6:

The area for this shape would therefore be the area of the triangle plus the area of the rectangle:

$\large \Delta x = v_o \Delta t + \frac {1}{2}\Delta v \Delta t$

Which, after distributing the second term for $v-v_o$ and combining like terms, simplifies to :

$\large \Delta x = \frac{1}{2}\left(v_f + v_i \right)\Delta t$

Which I’ll call Equation 2, which also gets the red encirclement.
From there we can algebraically combine Equations 1 and 2, to get equations which are easier to use in common situations.
If you know acceleration, but not final velocity, rearrange Equation 1 so it is explicit for $\Delta v$ and substitute that into the unsimplified form of  Equation 2.  That looks something like this:
$\Delta v=\overline {a}\Delta t$
$\large \Delta x = v_o \Delta t + \frac {1}{2}\Delta v \Delta t$
$\large \Delta x=v_o\Delta t+\frac{1}{2}\left(\overline{a}\Delta t\right)\Delta t$
$\large \Delta x = v_o \Delta t + \frac {1}{2}\overline {a}\left( \Delta t \right)^2$
Which is circled in red and called Equation 3.
If the time interval is not known, you can rearrange Equation 1, so that it is explicit for $\Delta t$ and then substitute into Equation 2.
$\large \Delta t = \frac{\Delta v}{\overline{a}}$
$\large \Delta x = \frac{1}{2}\left(v_f + v_i \right)\Delta t$
$\large \Delta x =\frac{1}{2}\left(v_i +v_f \right)\left(\frac{v_f-v_i}{\overline{a}}\right)$
$\large \Delta x=\frac{1}{2\overline{a}}\left(v_f +v_i\right)\left(v_f – v_i\right)$
$\large \Delta x = \frac{v_f^2 – v_i^2}{2 \overline{a}}$
Which would be Equation 4, and also get the red box.  With that, you have the four equations of kinematics:
$v_f = v_i + \overline{a}\Delta t$
$\large \Delta x = \frac{1}{2}\left(v_f + v_i \right)\Delta t$
$\large \Delta x = v_o \Delta t + \frac {1}{2}\overline {a}\left( \Delta t \right)^2$
$\large \Delta x = \frac{v_f^2 – v_i^2}{2 \overline{a}}$