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(Largely containing a mind-dump to myselves: past, present, and future)
Current favorite quote: "Supposedly smart people are weirdly ignorant of Bayes' Rule." William B Vogt, 2010

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.


Posted on 2010-11-15 by Jach

Tags: math, physics

Permalink: http://www.thejach.com/view/id/142

Trackback URL: http://www.thejach.com/view/2010/11/deriving_the_kinematic_equations

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Larry 15 September 2012 03:09:17 AM
I don't understand where the denominator "2" comes from in the 3rd kinematic equation. It's probably something simple that I'm missing. Perhaps you can clarify this for me.
Jach 15 September 2012 03:28:52 PM
Hey Larry. The third equation says that the final position of an object is the average velocity multiplied by time, plus a starting position. Equivalently, the average velocity is equal to the total distance traveled divided by time. (m/s).

Averages in general are calculated by adding up the relevant data points and dividing by the total number of data points that were added. Since we're only adding up the final velocity and the initial velocity, we divide by 2. That result is the average, it's another way of calculating it besides change in distance over change in time.

----

There are some assumptions here that I glossed over when I wrote this post... Why pick the initial velocity and final velocity to use as sample points for the average, as opposed to some other velocities in the time interval of interest? It's arbitrary, unless I assume that acceleration is constant. That's an okay assumption because in the fourth equation, the acceleration variable "a" does not depend on time so it is constant. For the problems where kinematics are useful (like cannon shots and free-falling objects), it's also not a bad assumption. If I drop a ball, it starts at 0 and ends at say 10 m/s, the average speed was 5 m/s. (Note we have to adjust the reference frame for time to be the moment before collision, or else the average would be 0 m/s and that's not true.)

In reality, if I wanted my true average velocity I would have to have an equation for velocity in terms of time and integrate over all the instantaneous velocities in a time interval, and divide by a factor of the final time minus the initial time. (And you can always get an equation for velocity if you have an equation for position or acceleration that depend on time.)

----

An interesting thing to think about is how different speeds affect the average. Here's a fun problem: A car travels a distance of 60 miles at an average speed of 30 mph. How fast would the car have to travel the same 60 mile distance home to average 60 mph over the entire trip?

The answer is it can't. In order to average 60 mph over the entire trip, the car would need to travel the 120 miles within 2 hours. (Distance over time.) Any longer and it is below average, any shorter and it is above average. Since it averaged 30 mph on the way there, that means it's already been traveling for 2 hours. Too late. The point is that periods of low speeds with destroy your overall average, and by the time you notice it may be too late to fix it.
Larry Schlanger 16 September 2012 02:16:43 AM
Thanks so much for your response. It was very helpful. I think it is fantastic you are willing to share your knowledge so graciously.

I am a physician, far removed from math and physics training but I am trying to relearn some of it for the purpose of helping my son get through it but also because I just love the stuff.

Any suggestions for a basic physics text?

Thanks again.
Jach 16 September 2012 08:51:04 AM
No problem, glad I could help.

For textbooks, the standard for high schools and undergraduates in college is still Physics for Scientists and Engineers, and there are camps of people who prefer the latest edition while other people prefer the 4th edition when Giancoli was the author. I don't have much preference for either in particular.

Khan Academy has become very popular over the past year and a half. The free videos they have cover a broad range of topics pretty well and in a short time period (here is the mechanics section). One of the neatest features is their Knowledge Map for math, where anyone can practice the subjects and let someone else (a "Coach") have access to the results. That is great for teachers or tutors because then they know exactly what areas a student is struggling with.

Walter Lewin's MIT class in 1999 is also good material, and he usually ends a class with some sort of experiment. I don't know any book of experiments, but doing the fun ones help cement the understanding.

If you want The Best Physics Text Ever (in my opinion), check out Richard Feynman's Feynman's Lectures on Physics volumes 1, 2, and 3. (Though 1 is all you need if you just want the Newtonian mechanics.) It doesn't have problem sets to solve and it's not exactly basic (you need to be pretty comfortable with calculus), but it's still excellent.

If you want a digital copy to check out either Feynman's Lectures or the Giancoli/non-Giancoli book I can email you a link--libraries should have the latter two but they don't always have Feynman's biggest work.

----

If you or your son are ever interested in electricity and magnetism, all of the above resources cover the subject but I think Introduction to Electrodynamics by Griffith along with Div, Grad, Curl, and All That by Schey as an optional supplement are a better combination. Outside electrodynamics in general and into computer engineering, I'd recommend Digital Electronics: A Practical Approach by Kleitz because even though it's pretty basic it's also fun to follow along and build interesting circuits out of regular wires, chips, and a breadboard.
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