2007-09-24

Today's nonsense
- Meg's still missing



Today we have a new assignment in two days. "So! Let us begin, shall we?" Homework: pg 76-77, #31-34,36-40.

ex 1) A particle's position coordinates (so we can even use the same terminology if you wish) (x,y) are the following coordinates: (3,7) at t=0; (7 m, 12 m) at t=4 sec; and finally (13 m, 17 m) at t= 10 sec.

(A) Find the V_avg from t=0 to t=4.sec
(B) Find the V_avg from t=0 to t=10 sec

So we have to look back to yesterday's notes to find that VectorA_avg = (change in VectorV)/(change in t). On #32 they ask for the average velocity, so maybe they just want a number here and not the vector? So the average velocity is going to be change in position over change in time. So in this case we're not going to be using specific vectors ... right now for (A) they are just looking for a specific number. On the test this will be further specified. So this is going to be "4i + 5j" / 4 ==> i + (4/5)j. So the magnitude is sqrt(41)/4 = 1.601.


So this next one ... oh my, I don't want to do this one, it's a thought problem. It's one of those where when the ball is at the maximum height what's the acceleration, things that are just testing to see if you know what's going on.

2) At T=0, a bug located at the origin that has a velocity of 24 m/sec at theta = 30 degrees. At T= 3 sec the bug's at x=60 m and y = 100 m with velocity 30 m/sec at theta = 60 degrees. So we have to find two things: (A) the average velocity and (B) the average acceleration (because it changed velocities). Find the velocity and acceleration vectors. So what's the average velocity vector? So that's the change in VectorR over change in T. So that's 60i + 100j (because we start at 100) and all divided by 3, so we can write that as 20i + (100/3)j so the magnitude of this is going to be sqrt(400 + (100/3)^2)--- but we're just finding the vector so we stop at 20i+(100/3)j.

Now we're going to find the average acceleration vector. So we do VectorA_avg = change in VectorV / change in T. So the starting velocity vector (V_1) is what and what's the ending velocity vector (V_2)? The V_1 is 24 m/s at 30, so you have to find your i and j-- you know that your i=12sqrt(3) and j=12, so you're going to be using a 30-60-90 triangle. So the V_1 is going to be 20.785i + 12j. The 30-60-90 triangle: the short leg is (1/2)hypotanuse, and the other one is .... right. So what about V_2 = 15i + 25.981j (where i and j are your unit vectors). The longer side is the sqrt(3) * hypotanuse, and the short side is (1/2) * hypotanuse, in a 30-60-90 triangle.
- So now we can do the delta to get the average acceleration vector. VectorA_avg = -5.785i + 13.981j all over 3 = -1.928i + 4.660j.


In cal 2, it's like the second and third test of next semester, where we have vectors, and one will not even have calculus in it, and that's next semester.

There are a couple problems. Mary and Robert. We might do #40 together tomorrow. So they are going to rendevouz on Lake Michigan, but we don't have time to make one up today like this, and we can maybe xerox the pages.