 Okay, so this is part two of the problem I started online, and so it has to do with this slide at the fair. I think I figured out what the problem was with my last calculation, but I'm not going to say what. So this question says, here's your slide, and I'm going to model it as a Bing Street, even though it's not. And then it has some level part like that. And so the question is, how long should that be? So you need the data from the first part, and I said 32 meters per second, and that's questionable. So let's just use something I got from the video. I actually found that the initial velocity was around 7 meters per second. So let me use that. And so the question is, how far should you design this such that people will stop and not crash off into the ground, which could be anything like asphalt or something? Okay, so I'm going to do this as a kinematics problem, a force and acceleration problem. It could be done as work energy. But again, let's just do what we did before. First we need to choose our x and y axis. If I have the slider on this part, then they'll be moving that way and accelerating this way. So it'll be all in the x direction, and that's good. Free body diagram, here's the slider, and then we have the gravitational force, we have the normal force, and then we have the frictional force. So it's very similar to the previous problem. We'll use the same model for friction that the magnitude of the friction force is in coefficient, which I found before, UK was about 0.31 times the normal force. So in the vertical direction in the y direction, there's no acceleration, so f net y equals zero is n minus mg. Okay, so in that case, now these are the y components of the forces, that's why there's a negative sign on the minus mg. And so here you can see that n does equal mg, so that's a little bit different. So now I can find the friction force, the magnitude of the friction force is going to be UK mg. And then in the x direction, f net x equals max, call that my positive x direction, so it's going to be max equals negative mu K mg. And the negative sign here, if you look, that's my positive x direction, so that's pointing in the negative x direction, I need that negative sign there, and the mass is cancelled. Okay, so I have the acceleration, which I also actually is weird because I actually used this, I measured the acceleration to get the coefficient of friction in the post, so I didn't really have to do that. I could have just started with the acceleration, okay, sorry. Okay, so now how long does it need to be? Well, if you want to use kinematic equations, you could start with this, v final squared equals v initial squared plus 2A x minus x naught. I could use that, and then I could just solve for x minus x naught is going to be the distance s and just solve for that. I think I have a little bit of time then I would like to use, not use that equation because no one really knows where that, forgets where that comes from. So let's just do this with basic things. Let's say acceleration is the change in velocity, and this is just all in the x direction, so I'm going to leave off, really this should be A x v x over delta t. And also I can say v x average is delta x over delta t. I can also say the x average is going to be the final which I'll just call v plus the initial over 2, that's the definition of average, and actually that only works if the velocity is changing at a constant rate, so you can't always say that, but that would be true if your acceleration is constant. Okay, so right here I want to find x, delta x which in this case is just s, but I don't know delta t, but I do know delta t from up here. So I can say delta t equals, I'll just write this as v minus v naught over A x, and so putting that together with this, I'll say s over v minus v naught A x, because I divide by that, equals v plus v naught over 2. And now I want to solve for s, I'll multiply both sides by v minus v naught over A x, and I get s equals v plus v naught v minus v naught over 2 A x, and this is a, when I have v plus v naught times v minus v naught, I get v squared minus v naught squared over 2 A x, which is if I solve this for s, I get the same thing, so it's the same thing. So now we can just put in our values, my final velocity is going to be 0, I want them to stop. My initial velocity I'll use is 7, and the acceleration is this value right here, negative. Okay, so let me put that in. So 0 minus initial velocity squared, 7 meters per second, quantity squared, over 2, negative 2 times A x, which is 0.31 times 9.8, let's just write it as meters per second squared, it's the same thing. And now let's just first check the units, on the top I have meters squared per second squared divided by meters per second squared, second squared is canceled and one of the meters cancel, I do get meters. And so I get, and it's going to be a positive number too, because these are going to cancel. Let me just put this in my handy-dandy calculator here, one second. Okay, so I have 7 squared, 3, 1 times 9.8, and I get 8 meters. Wow, that's two fails for me in a row, 8 meters, what 24, I guess that's not too bad, 8 meters isn't completely unreasonable, if you look at those tracks, I mean they're pretty long, the level part. And in fact, I saw people going over the end, I saw them sliding and they didn't even come off the end. So I don't think that's too unrealistic and I'm okay with that, are you okay? Let's all be okay with that, okay.