3/27/11

Soooo, time for another blog. This time I'm going to go over PVA, which we all know stands for Position, Velocity, and Acceleration. So, to go over the basics: If you're given a position function and they're asking for velocity, you're going to take the derivative of that equation (i.e...for "PVA", DERIVATIVE TO THE RIGHT; INTEGRAL TO THE LEFT)..If you're given position and they ask you for acceleration, you take the derivative twice. If you're given velocity and they're asking for the position (or distance), then you're going to integrate the equation they gave you and add your +c to that. With that, here are some examples:

Ex. 1) At t=0 a particle starts at rest and moves along a line in such a way that at time t its acceleration is 24t^2 feet per second per second. Through how many feet does the particle move during thye first 2 seconds?
*Okay first let's identify what we're given to work with in the problem:
They tell you 24t^2 is the acceleration; they're asking for how many feet the particle moves during the first 2 seconds, so that means you're going to integrate the equation TWICE because "the FIRST 2 seconds" implies that they want you to use the position function
*So integrating 24t^2 for the first time you get 8t^3; then integrating it a 2nd time you get 2t^4
*Now you plug in you time value (2):
2(2)^4
2(16)
=32

Ex. 2) A particle moves along the x-axis so that its position at time t is given by x(t)=t^2-6t+5. For what value of t is the velocity of the particle zero?
*So for this problem we are given the position function and we're asked to find the velocity at zero; All that means is to derive the position function to get velocity and then set it equal to zero and solve for x
*So deriving the function you should get: 2t-6
*Set it equal to zero and solve for x:
2t-6=0
2t=6
t=3

Ex. 3) The maximum acceleration attained on the interval 0*So for this problem they're asking for maximum acceleration. So first you're going to have to derive the velocity function they gave you to get acceleration.
*Doing that you get: 3t^2-6t+12
*Since they say MAXIMUM, you're going to also have to do the 1st derivative test
*So deriving the equation you just got, you should now have: 6t-6
*1st derivative test says to set the derivative = 0: 6t-6=0...and solve for t
t = 1
*Set up intervals: (*they gave you the interval [0,3] so you MUST use that!)
(0,1)u(1,3)
*plug in values on both intervals (1st interval is negative; 2nd is positive, which means there is a max at x=1)
*Alsoooo, since you were given an interval in the problem, you MUST check your endpoints:
f(0)=3(0)^2-6(0)+12 = 12
f(1)=9
f(3)=21
*And whichever number you get once you've plugged in is your maximum value
*So your maximum acceleration is 21