Hey, guys. Let's check out this problem together. We've got a subway train and it's going to start from rest and accelerates, then it's going to travel at constant speed for some time, and then finally, it's going to slow down until it stops at the next station. So there's a bunch of moving parts here with multiple accelerations. So we're going to solve this like any other problem with multiple parts with accelerations. The first thing we have to do is just draw the diagram and list all the variables for all of the intervals that we have. So the first part, which is where the train is accelerating, I'll call it A to B. Then it's going to travel at constant speed. So from B to C, it's going to be constant speed. And then finally, it's going to stop as it approaches the next station. So this is actually going to be from C to D and it's going to be stopping here. So we've got a bunch of variables to keep track of. Let's just go ahead and list them out. So from A to B, we've got Δx from A to B, which we know, which we actually don't know. We don't know Δx from A to B. That's the distance in the first part. We've got the velocity starting from rest, so this is just going to be 0. And let's see, we, so we know this is 0. We don't know the final velocity, but we do know the acceleration and the time. So I do know my acceleration from A to B and T from A to B. This is just equal to 1.5 and this is going to be 14 seconds. So now we just have to figure out whether the 1.5 is positive or negative. Well, it's going to start from rest and it's going to accelerate, which means it's going to speed up in the right direction, so that's going to be positive. So it's going to be 1.5 positive. Okay. For the second part, now we've got constant speed for 60 seconds. So we've got Δx from B to C, we don't know what that is. Vb, we don't know what that is either because we don't know what it is from the first part. VC, don't know what that is. Acceleration from B to C is going to be, well, it says constant speed. So this constant speed here actually means the acceleration is equal to 0 for this part. But we do also know that the time for part B is 60 seconds. So whatever this velocity ends up being, VB and VC, they're actually going to be the same thing because there's no acceleration. Alright. And so finally, now we have the stopping part. So we have Δx from C to D. We don't know what that is. So we have Δx from C to D, we don't know what that is. VC, we don't know what that is. Vd, that's just going to be when it stops at the next station right here. So this is going to be 0. And then finally, the acceleration from C to D, well, it's going to slow down at 3.5 meters per second squared. So first, it accelerated then it went to constant speed and then it's going to slow down which means that this acceleration is going to be negative 3.5. And then we got T from C to D and we don't know what that is either. So we've got all these variables here. That's the first step. Now we just have to figure out what the target variable is in our problem. We're trying to figure out what's the total distance that the train covers. So remember, we have all of the different parts of this motion here, but we also have the whole entire interval. And this whole entire interval here, there's a distance, Δx from A to D. Right? It's the total distance that you take from here all the way out to here. That's what we're trying to find. So Δx from A to D is our target variable. How do we figure that out? Well, we don't have an equation for Δx from A to D using our motion equations, but we can figure it out by piecing together all of the different distances in each of the parts. So, right, if this was 10, 20, and 30, you just add them all together and that would be your total distance. So this guy is actually going to be Δx from A to B, plus Δx from B to C, plus Δx from C to D. So we're going to have a bunch of parts to figure out. And once we figure out all of those parts, we can just add them together for a grand total. So we've got to figure out what all of these parts are because we actually don't know what any of them are off the bat. So let's figure that out. So let's see. I'm going to look for Δx from A to B. So this is going to be my target variable in this piece right here. So that's the second step, which is where we identify the known and the target variables. Now we have to pick a UAM equation, one of our motion equations that doesn't have the ignored variable. Notice how I have 3 out of 5 variables, 1, 2, 3, and which means the one I'm ignoring is actually going to be the final velocity, my VB. So that means I just need to pick an equation that does not have this VB in it. And if you go ahead and look for that, that's actually going to be Well, let's see. We have VB here. This is final velocity, final velocity and this is also going to be final velocity. So we're actually going to pick equation number 3. So this is going to say that my Δx from A to B, which is actually my target variable, which is great because this is only going to be on the left side of the equation, is going to be my initial velocity times time, but we know this is going to be 0 because our VA is equal to 0, plus one half acceleration which is 1.5 times T squared. So this is going to be my 14 seconds squared. So I can just actually plug that straight into my calculator and my Δx from A to B is going to be 147 meters. So that's one of my variables, 147. So I'm one step closer to figuring out the total distance. So that's good. Now, I just need to figure out the next part which is the displacement from B to C. So now this is going to be my target variable here, Δx from B to C. So we'll go and focus on that. So how do I figure this out? Well, my Δx from B to C. So which equations am I going to use? Well, remember that we only use the 4 UAM equations whenever the acceleration is constant and it's not 0. When the acceleration is equal to 0, then all of our equations simplify and we only just use 1, which is V equals Δx over Δt. So if we're trying to figure out Δx from B to C, we're just going to use the constant velocity formulas. My VB to VC is equal to Δx from B to C over TB to C, which means that Δx from B to C is just going to be my velocity times the time. Now I have the velocity, or sorry, I have the time, which is 60 seconds here. So I have that. All I have to do is just figure out what the velocity is from B to C. And notice how it's actually going to be the same velocity. It's going to be either VB or VC because it's going to be a constant velocity. Right? So whatever I figure out for VB is going to be the same thing for VC. Now we just have to go and figure that out. How do we do that? Well, notice how if I don't have either of them, I'm going to have to go to a different interval and figure that out. Which other interval also has VB? Well, these either well, actually, there's only one choice. It's going to be the first part right here. So what we could do is we can actually use the first interval, because now we know 4 out of the 5 variables, to figure out what my velocity is that I ignored in the first part. So that's what we're going to go ahead and do. So now what I can do is I can go back over here and I can say, well, now I want to figure out what my velocity B is. So how do I do that? Well, now I'm going to use my 3 out of 5 variables for this part here. And if I'm trying to figure out the final velocity, I can just use the first equation. That's the easiest one. So I'm going to use equation number 1 to figure out the final velocity. So VB is equal to VA plus A times T. So I know I'm going from rest, so this is 0. And then my acceleration is 1.5 and my time is 14, so I get 21 meters per second. So that means that I know that this is 21 and this is also 21, which means now I can go ahead and use my formulas. Now I have the velocity and I have the time which means I can figure out Δx. So this is just going to be 21 times my TBC which is 60. And if you work this out, you're going to get 1260. So notice how sometimes you're going to have to stop and you're going to have to go to other parts in order to figure out some variables over there and bring it back to the equation that you're trying to look for. So that's the second piece. Think about it like a piece of a puzzle. I've got the second piece of the puzzle, so 1260. So now I'm just going to have to figure out the last one. So the last one is Δx from C to D. So I've got to figure out this guy over here. Notice how also I figured out another variable when I did this. So remember how VC was an unknown? Well, we just figured out that VC was 21. Here it traveled at 21 meters per second, so that means it ended right here and now it's 21 meters per second over here and then eventually it's going to stop. So now we're looking for the Δx from C to D. We just go ahead and look at our UAM equations. We know there is an acceleration here, so we are going to use these variables, which means we do need 3 out of 5. I've got 3 out of 5, 1, 2, and negative 3.5, which means that my T is going to be my ignored variable over here. And so now which equation we're going to use? Well, the one that T is ignored in is going to be equation number 2. So this is the one we're using for 2, for Δx from C to D. So this is just saying that my final velocity VD squared is equal to my initial velocity VC squared plus 2 times A from C to D times Δx from C to D. So this is actually what I'm looking for here, my Δx. So now you just work everything else out, you plug everything else in. So we know my VD is going to be 0, my VC is going to be 21, and then 2 times the acceleration. Well, the acceleration is going to be negative 3.5, remember the negative sign, and then times Δx from C to D. So now, if I square this, I want to actually get and move to the other side. If I square this, move to the other side, you're going to get negative 441 equals and then you multiply these things out, you're going to get negative 7 times Δx from CD. So now, my Δx from C to D is just Well, the negatives will cancel and you're just going to get 441 over 7 which is equal to 63 meters. And so this is the final piece of the puzzle, the final displacement. So now you just go ahead here and put them all back together again. So 63. I can say that Δx from A to D, the total distance traveled of all of the parts is going to be 147 + 1260 + 63. You add all these things together and what you get is you get 1470 meters and that's your final answer. Alright, guys. So again, a lot of moving parts. That's why it's really important to keep organized in these problems, because otherwise you're just going to get totally lost when you're trying to figure these out. Anyway, let me know if you have any questions. That's it for now.
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