Continuing from yesterday's notes on acceleration, velocity, speed and vector quantities and scalar quantities and so on. The Dr. Pepper group has been acting up.
<br /><br />
<h1>Motion w/ constant acceleration</h1>
v(t) = integral of a * dt = at + c = at + v_0<br />
<br />In the above case, the "a" variable is a constant, so you have to evaluate it as such. This is why this is known as <u>constant</u> acceleration. When do you most often have constant acceleration? Free-falling is when you have constant acceleration, until you hit the terminal velocity. In this case, "c" is an initial velocity. <br /><br />
<br />
v(t) = at + v_0<br />
<br />
x(t) = integral of v(t) dt = integral of (at + v_0)dt = (1/2)at^2 + v_{0}t + c<br />
So when you take the integral of velocity, you get the position function. What is that value of "c"?<br /><br />
x(t) = (1/2)at^{2} + v_{0}t + x_{0}
<br /><br />
Using those two equations, find a single equation in terms of x, x_{0}, v, v_{0}, a. You can always start with acceleration and integrate to get the velocity, and then integrate again to get that above x(t) equation. Note that in this list of variables, there's no "t". So you want to substitute.<br /><br />

So how would you begin with that task? <br /><br />
t = (v - v_{0}) / a = (delta v)/a<br />
And the second equation becomes: x = (1/2)a(v - v_{0})^2 + v_{0}( (v - v_{0}) / a) + x_{0}
<br /><br /><br />
This notation is not quite LaTeX. Need to get the preview-latex module installed for visual emacs. Use this file with the written notebook notes.<br /><br /><br />

All of these equations are related. You can get all of these three equations by taking the integral of acceleration.<br /><br />
 Apparently Lockhart's final was "make a Rube Goldberg machine." 
 <br /><br />

<h1>For objects falling</h1>
For objects falling freely due to gravity:<br />
g = 9.81 m/s^2 = 32.2 ft/s^2
<br /><br />
The Tipler book keeps the negative sign with "g" into the equation itself rather than with the variable. The negative would indicate that something is falling down. You could also say that going towards the earth is positive, like when you have to wonder how far something falls and so you would want to call falling down as positive etc. We may "g" negative from time to time.<br /><br />

<h1>Interpreting graphs</h1><br />
<ul>
<li> x(t) also known as displacement <u>vs</u> time. What isi t about this graph? What can you find out besides deplacement? You can see that slope is velocity.
	<ul>
	<li>Slope is the velocity.
		<ul>
		<li>Velocity is increasing when graph is concave up.</li>
		<li>Velocity is decreasing when graph is concave down.</li>
		<li>Velocity is zero? Velocity is zero when the slope is zero, so it's the maxs/mins of the x(t) graph.</li>
		<li>Velocity is a maximum or minimum? At inflection points, which are <b>maxs/mins of v(t) graph</b></li>
		</ul>
	</li>
	<li>Acceleration
		<ul>
			<li>Acceleration is positive when graph is concave up.</li>
			<li>Acceleration is negative when graph is concave down.</li>
			<li>Acceleration is zero when it's (1) constant velocity, or at (2) an inflection point (the point where it changes concavity on the original graph). Or when the graph has a constant slope (i.e. is linear).<br />
		</ul>
	<li>
	</ul>
</li>
<li> Velocity versus time
	<ul>
		<li>Slope = acceleration</li>
		<li>Acceleration increasing for concave up</li>
		<li>Acceleration decreasing for concave down</li>
		<li>Velocity is constant when the slope is zero.</li>
		<li>What is the area under the curve of a velocity-versus-time graph? The area under the curve here is your <u>displacement</u>- change in position.
			integral of v(t)dt from a to b = x(t) from a to b
			</li>
		
		<li></li>
	</ul>
</li>
</ul>