Deriving the Kinematic Equations

The Kinematic equations are the bread and butter equations of Newtonian projectile motion, and are quite useful for games. There are four of them, but you just need to remember two very intuitive ones and you can derive the other two from there. For one derivation, we need Calculus, for the other, it's straight Algebra. But don't be afraid, these are trivial! I'm mostly doing this to refresh my memory and it may help someone.

First, some definitions:

[math]\begin{eqnarray}
X &=& position,\ distance\ traveled \\
V &=& velocity \\
a &=& acceleration \\
t &=& time \\
?_0 &=& initial\ value\ of\ whatever\ ?\ is \\
?_f &=& final \ value\ of\ whatever\ ?\ is
\end{eqnarray}
[/math]

Pretty straightforward. So, the First Kinematic Equation? Well, what is velocity anyway? If I know I'm accelerating forward at 10 m/s^2, and I travel for 2 seconds, how fast am I going at the end? 10*2 = 20m/s. If I was going 5 m/s at first, then I'm now going 20+5=25m/s. So velocity is a function of acceleration and time plus an initial velocity:

[math]V_f = a*t + V_0[/math]

Indeed, if we look at the units, they match up:

[math]V_f (\frac{m}{s}) = a (\frac{m}{s^2}) * t (s) + V_0 (\frac{m}{s})[/math]

The seconds in the time factor cancel one of the seconds in the acceleration factor, leaving a sum of m/s.

Now let's integrate with respect to time!

[math]\int{V_f dt} = \int{(a*t + V_0)dt} = \frac{1}{2}*a*t^2 + V_0*t + C[/math]

Convince yourself that when we integrate velocity over time, we get back position, and therefore the C constant is just the initial position:

[math]X_f = \frac{1}{2}*a*t^2 + V_0*t + X_0[/math]

And that is the Second Kinematic Equation.

The Third Kinematic Equation is another intuitive one. What else is position, if we don't have information about acceleration, but only velocity? If I'm traveling at 60 mph, and I travel for two hours, how far have I gone? 120 miles. (Plus an initial position, of course.) But in the real world, I can't just jump to 60 mph instantly, and there may be stop signs, slow downs, speed ups when no cops are around... But if I know my average speed is 60mph, then the above holds. Position is thus a function of average velocity and time, the third equation:

[math]X_f = \frac{V_f + V_0}{2} * t + X_0[/math]

And looking at units:

[math]X_f (m) = \frac{V_f + V_0}{2} (\frac{m}{s}) * t (s) + X_0 (m)[/math]

The seconds in time cancel out the seconds in velocity, giving a sum of meters.

To get the final Kinematic equation, it's straight algebra from here. Let's take the third equation, solve for t, and plug it in to the first equation. Why not? If we're careful we can get an equation that will let us know our velocity from simply an acceleration and position alone, no need for time.

[math]\begin{align*}
Solve\ for\ t: \\
\frac{V_f + V_0}{2} * t &= X_f - X_0 \\
t &= \frac{2 * (X_f - X_0)}{V_f + V_0} \\
\end{align*}
[/math][math]
\begin{align*}
Now\ plug\ it\ in: \\
V_f &= a*t + V_0 \\
&= \frac{2*a*(X_f - X_0)}{V_f + V_0} + V_0
\end{align*}
[/math][math]\begin{align*}
Multiply\ both\ sides\ by\ (V_f + V_0): \\
V_f^2 + V_f*V_i = (V_f + V_0) * (\frac{2*a*(X_f - X_0)}{V_f + V_0} + V_0) \\
\ \ \ = V_0^2 + \frac{V_0*2*a*(X_f - X_0)}{V_f + V_0} + V_f*V_0 + \frac{V_f*2*a*(X_f - X_0)}{V_f + V_0} \\
\end{align*}
[/math][math]\begin{align*}
V_f^2 &= \frac{2*a*(X_f - X_0) * (V_f + V_0)}{V_f + V_0} + V_0^2 \\
&= 2*a*(X_f - X_0) + V_0^2
\end{align*}
[/math]

And that's the Fourth Kinematic Equation.

Simple, right?

Edit: There is also a second derivation for the fourth equation, following from conservation of energy. Recall that:

[math]Kinetic\ Energy = \frac{1}{2} * m * V^2 [/math]

and:

[math]Potential\ Energy = m*g*h[/math]

where m is mass, V is velocity, g is acceleration due to gravity, h is positional height. Recall the law for Conservation of Energy:

[math]PE_0 + KE_0 = PE_f + KE_f[/math]

So, by simple algebra:

[math]m_0*g_0*h_0 + \frac{1}{2}*m_0*V_0^2 = m_f*g_f*h_f + \frac{1}{2}*m_f*V_f^2[/math]

We assume that mass is constant (so we can divide it out) as well as the gravitational acceleration:

[math]\begin{align*}
g*h_0 + \frac{1}{2}*V_0^2 &= g*h_f + \frac{1}{2}*V_f^2 \\
Multiply\ both\ sides\ by\ two: \\
2*g*h_0 + V_0^2 &= 2*g*h_f + V_f^2 \\
V_f^2 &= 2*g*h_0 - 2*g*h_f + V_0^2 \\
&= 2*g*(h_0 - h_f) + V_0^2
\end{align*}[/math]

So this form is essentially the same as the above.

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Posted on 2010-11-15 by Jach

Tags: math, physics

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