 Okay, so this is the Thursday lamb, section six, I'm going to go over the test. Okay, so the first question said, I mean you've got to read these questions carefully. It's not saying just what's the launch velocity, we're launching a ball in three different ways. It's saying which one's the best also, okay. So you kind of have to read the question correctly. For the first case, the ball starts straight up and comes back down and the lab person records a time of 0.6 seconds. So we'll have t equals 0.62 seconds. And so how would you calculate the velocity? Well, what's the time to get to the highest point? It's half that. And then you can say v equals v naught minus gt and the final velocity is zero. And so the initial velocity of solving for that v zero equals gt. So it's going to be 9.8 times 0.31 seconds. And that gives me 3.03 meters per second. The next method, you shoot it horizontally and you measure the two distances. This is a problem that we did in class. So like that. So I measure y. I measure x. I can get the time from the y coordinate with y equals 1 half gt squared. And then I can get the, since I shot it horizontally, v zero is going to be x over t and solve t from there. And when you do that, you get 2.4 meters per second. And the last case is the same idea of the different measurements where you measure x and time of 0.31 seconds. And you get the velocity of just v equals x over t and you get 2.42 seconds, meters per second. So the question is, which one is the best? So from the lab, this one is the best value. Because how hard is it to measure y and x? It's not hard at all. You'd have some uncertainty in there and some fluctuations. But trying to time 0.6 seconds with stopwatch or 0.31 seconds with stopwatch, as you did that in class, you're going to get widely different answers. So that's not a very reliable answer. Even though this one is close to that, a lot of people said, oh, they're close. Therefore it doesn't matter. This one just happened to be close by chance. You click in that stopwatch and you make it a value that gives an OK value, but it doesn't mean that it's a good method. This is a much better method. OK, next question. Here I gave you some mass and volume data. And I want you to find the density from a graph. So density is mass over volume. So I could write this as mass equals density times volume. So that's a function like y equals slope times horizontal. So if I put this as mass and volume and I get data and I plot it, I can find the slope of this line. And if I use the slope in the units in grams and cubic centimeters, you get a slope around 2.7 grams per centimeter cubed. And since I'm applying mass or some volume, that would be the density. That's all there is. That's it. Just practice in making a graph and finding the slope and relating that slope to something. The slope does have units. It's changing this over changing that. So it would be units of grams per cubic centimeter. Or you could change it to kilograms and cubic meters. It doesn't matter. The last one has a car and a ball and there are one's constant velocity and one's accelerating on a ramp. And so the first question is make a sketch of when they would meet. So here's the car. It's just going to have a slope of 1.2 meters per second. But the sketch, you don't have to put that on there. Now the ball starts out at rest and it's going to increase in speed and so it's going to look like this. And so at some point after when they started, they'll meet again. And that's the point we want to find. So the next question is where and when does it catch up? So I can write down an expression for this car. I'll call that x1. It's just going to be 1.2 meters per second times t. That's your constant velocity expression for that. So t equals zero, it's a zero. Where the ball is going to be x2 equals, it started at x equals zero and it started with initial velocity of zero so it's just going to be 1.5 times 1.2 meters per second squared. I gave those two the same numerical value just for no reason, just for fun, t squared. Now I want to say when are these two things equal, x1 equals x2. So I get 1.2t equals 1.5, 1.2t squared, I'm sorry, yeah, the one point. So I get 1 equals 1.5t, t equals two seconds. When does that, so that's when they meet, where do they meet? Well now I can put that same time into either one of these functions and I get x1 equals 1.2 times 2, 2.4 meters. What about this, calculate the time that the car has the same speed as the ball? Well, the speed of the ball, the car is constant. What about the speed of the ball? V equals v0 plus at, that's the expression for the speed. So I want to say, it starts at rest, 0, so this is 1.2 meters per second equals 1.2 meters per second squared, t, so t equals one second at the same speed. So here they don't have the same speed, they have the same speed down here.