2007-09-06
Today's nonsense:
- All "Megan"s missing
- Two volleyball players missing
- Dakota will be missing tomorrow
There was a question on the homework, #37. You start at 48.3 km/h and go up to 80.5 km/h in 3.7 seconds. So obviously they want the average acceleration in m/s^2, so average acceleration is going to be the change in velocity divided by the time. So that's 32.2 km/h per 3.7 seconds.
Test hint: the Joe&Sally problem from the notes.
And now for some examples that you will need for the test: at the moment we have learned everything we need for the test, but now we're just trying some different approaches to solve basically the same problems.
Ch. 2 #69- would have been a problem on homework but is obviously not so anymore. A stone is thrown vertically from a 200 m cliff. So starting position = 200. During the last half second of its flight, it travels 45 meters. Find the initial velocity of the stone- somebody threw the stone vertically. So was the stone thrown vertically up? Or was it thrown vertically down? Take the fact that it travels 45 m / 0.5 sec, that's your average velocity. So average velocity is going to be 45 m / 0.5 sec or 90 m/sec. And we are still trying to find initial velocity. What's another way to find average velocity over a time period? .... Anyway, eventually you are able to get V_final, so now you're able to use something like (V_f)^2 = (V_0)^2 + 2a(change in x).
Your delta x (or delta s, whatever) is -200. Because it's S-S_0, and S when you hit the ground is 0, and S_0 was 200.
So is s = (1/2)aT^2 + V_initial * T + S_0 always true? No, only for constant acceleration. See the 2007-08-29 lecture. Otherwise you'd have to integrate the given acceleration.
Chapter 2 #71. A bus accelerates at a = 1.5 m/s^2 from rest for 12 seconds. S_0 = 0, V_0 = 0, and then afterwards it goes at a constant speed for 25 seconds, and then it slows to a stop at a = -1.5 m/s^2.
Acceleration is positive at first, so it's going to start with concave up, if you want to graph it. But you could graph velocity.