Poistion, velocity and acceleration

displacement- change in position
dx is change in position

dx = x_2 - x_1
Where you ended up minus where you started. Can you have a negative displacement? Displacement can be negative because negative tells you direction. Four meters? Well, four meters which direction? And right now it's only one dimensional, two dimensional w/ vectors is later.

Displacement is a vector. Distance is a scalar. Velocity is a vector. Speed is a scalar.

(delta S is the same thing as delta X in displacement problems)

average velocity- velocity at the end minus the velocity at the beginning, all over the change in time. (dx / dt). Velocity goes with displacement. Is it the distance divided by time? Or is it displacement divided by time? In the case of average velocity, it's displacement divided by time since velocity is a vector and so thus displacement must be used since it is a vector.

400 meters around the track outside. If you go 400 meters in 60 seconds, your average velocity is going to be zero since it's displacement and you ended up where you started.

Now, with distance over time, that's average speed. Average velocity is different.

Average of the mean: V with a bar over it. 
Average velocity is going to be dx/dt, which is x_2 - x_1 all over t_2 - t_1.

Displacement, velocity, and acceleration are all vector values. They have a magnitude and direction. For example, distance and speed are scalar quantities. Which just means that they are the magnitude (number) part. 
	Displacement. 
	Do you remember when you were given a velocity function, and you integrate the velocity, what do you get with respect to the graph? You're getting the area under the curve. You were getting displacement. The curve could have been above or below the axis, and so when you add together the area below and above you get a final answer. Neglecting the negative values, you were able to get distance (take the absolute value of each term). 
	Displacement = the integral of v(t)dt from a to b. How did you find distance?
	Distance = the integral of |v(t)|dt from a to b. The change in the position function is displacement. 

So displacement is the integral of velocity. So distance is the integral of speed. The reason why this is so is because displacement has to tie into a vector quantitiy, and distance is only a scalar. In the case of distance as the integral of speed, you're getting the magnitude portion of the vector quantity.

Speed = |velocity|
Integrate the absolute value of velocity, you're integrating velocity. Velocity could be a function or just a number. Velocity is magnitude and direction. The magnitude is the speed. The absolute value drops the other component of the vector quantity in the case of |velocity|

How do you find average speed?
Average speed = (distance)/(time)

When you change something about the velocity, like the direction or the magnitude, then you're accelerating. So if you turn in a circle very quickly, you're changing your acceleration. This also goes to angular speed and angular acceleration later on.

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Instantaneous velocity- 
	v(t) = lim as (delta t)->0 of (delta x)/(delta t) = (dx)/(dt)
	NOTE: dx etc. is an approximation, "delta x" is an actual number
	What does the derivative tell us about the graph? 
	In this case it tells us the rate of change or slope. 
	v(t) = slope of the x vs. t graph

The magnitude of the instantaneous velocity is instantaneous speed. To get magnitude you take the absolute value, and then that's the number and that number is speed. 

Acceleration is change in velocity. Acceleration can be a negative number, and in that case it is deceleration. 

Average acceleration = (a with a bar over it) = (delta v) / (delta t) 
		     = (V_2 - V_1) / (T_2 - T_1)
		     = (V_f - V_i) / (T_f - T_i)

Instantaneous acceleration = lim as (delta t) -> 0  of (delta V / delta T) = 
			   = dv/dt
			   = the slope of v vs. t graph
	(the acceleration at a particular moment)

Average value of a function?




Position: x(t)
velocity: v(t) = change in position = derivative of position = x`(t) = (dx)/(dt)
Acceleration: a(t) = change in velocity = derivative of velocity = v'(t) = dv/dt
	= a(t) = v'(t) = x''(t) = dv/dt = d^{2}x / dt^{2}

What do you get if you integrate acceleration? You get velocity. 
integral of a(t)dt = v(t)
integral of a(t)dt from a to b = v(t) | (from a to b) == v(b) - v(a)
				= change in velocity
When you integrate, you are finding the change in the quantity.

integral of v(t)dt = a position function, like s(t) or x(t)
integral of v(t)dt from a to b = x(t) | (from a to b) = x(b)-x(a)