2/6/11

Soooo this week we started taking practice AP testing..well just the multipule choice part, and actually it is really hard. I know the information, but for some reason I just can't work the problems out..hopefully with practice I will get better at taking the test. Anyways, since we started taking the AP test we haven't learned anything new. I pretty much went over everything besides PVA, so i'll explain how to do that. PVA: position, velocity, and acceleration. The week before we took the AP test we learned about these functions and how to apply integrals with them. For instance, displacement or position is the integral of velocity; and distance is the absolute value of the integral of velocity. Most of the PVA questions will have more than one part to them. It's good to realize what the question is asking for (as far as distance/displacement).
Here's an example:

1.) Meg accelerates her car, giving it a velocity of v = t^1/2 - 2, in feet per second, at time t, in seconds after she started accelerating.

A: Find the time(s) at which v = 0
*Ask yourself, what is this problem looking for? They're asking for "t", so all you do is plug in 0 for v <--We know to plug in this because it tells us that at t v=0.
-So set the equation they gave you equal to zero and solve for t.
t^1/2 - 2 = 0
t^1/2 = 2 ..(t^1/2 is the same thing as the squareroot of t)
(t^1/2)^2 = 2^2
t=4

B: Find her net displacement for the time interval [1,9].
*Notice they give you the time interval [1,9]..which is going to be your bounds for your integral
*Also notice there asking for the displacement, you're going to take the integral of the velocity function with bounds 1,9
*So plugging into that formula you should get this:
S1,9 t^1/2-2 dt
Now taking the integral of that equation you should end up with:
2/3t^3/2 - 2t
Now use the Fundamental Theorem of Calculus to plug in 9 and 1:
2/3(9)^3/2 - 2(9) - [2/3(1)^3/2 - 2(1)]
= 4/3 ft
is displacement

C: Find the total distance she travels for the time interval [1,9].
*So now since they're asking for distance, all we're going to do is use the distance formula..which says to take the absolute value of the integral of velocity
*So plugging that into your calculator with bounds [1,9] you should end up with 4
*So 4 ft is your total distance

These problems are reallllly easy, you just have to follow the formulas that we already learned about distance and displacement, and also finding integrals. It's pretty much everything rolled into one. And that's alllll :)