1

concept

## Kinematics Equations

8m

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Hey, guys, you need to know how to solve motion problems involving in acceleration. So what I'm gonna do in this video is introduced you before equations of motion or sometimes called the Kinnah Matics equations that you absolutely need to know. But more importantly, I'm gonna show you when and how to use each one of these equations. So this is really important for you to learn practice and master because it sets the stage for the rest of physics. So let's check it out. So I remember for constant velocity, when the acceleration is equal to zero for motion problems, the only equation we could use is V equals Delta X over Delta T. We had another version of this equation that we called the position equation. But really, these two things were just two versions of the same equation. Well, a lot of times in physics you'll find that the acceleration is not zero. And in those cases you're gonna need form or equations called the Equations of motion or sometimes called cinematics equations or uniformly accelerated motion equations. Now, uniformly accelerated motion just means that the acceleration must be constant. And that's what you absolutely need to know about these equations. You can only use them when the accelerations constant. Now a lot of textbooks will show the derivations and the proofs and all that stuff, which you don't really need to know. But you should memorize them. So I have them listed in this table here. So we're gonna talk about each one of these equations. So it's over the first one, which is that VI equals vi not plus 80. This equation actually comes from an equation that were pretty familiar with. It comes from a acceleration average is equal to Delta V over Delta T. So you can manipulate this equation toe look like this in the same way we can manipulate the vehicles Delta X toe look like the position equation. It's the same idea. So we have the next one here, which is nothing really nothing special about it. V squared equals V not squared, plus to a Delta X. This third equation here have written in two different ways because you'll commonly see it written in those two different ways they look the exact same. The only difference is that this Delta X over here can sometimes remember it could be expanded and rewritten as X minus X initial, and they just move this thing to the other side. So again it means the same exact thing. You'll often see them sort of written interchangeably like that. So now the last one over here is Delta X equals V, not plus V over two times t. And this one has a little Astra's because in order to use this, I strongly recommend that you look at your ask your professors because some textbooks and professors might not allow you to use this. But basically what this equation says is this V Notch plus V is actually just the average of two velocities. So this is really saying here is Delta X is equal to the average times T, which we actually already know. Okay, so that's really it. So what? The important thing you need to know about these equations is that all of them have some combination of all of these five variables, and you need to know which equations have what variables in order to get the right answer. So let's go through each one of them really quickly. So we have V equals V not plus 80 So you have V, not A and T, but Delta X is missing some to write the little sad face there. Now we have V V, not A and and Delta X. So that means this is V V, not A and Delta X, but it's missing time. And now for these two equations for number three, whatever form you're using, this has Delta X has V, not T, and A So this is Delta X V, not T and A. But it's missing the final velocity. And now finally, we've got Delta X equals V not pose v times t s o. That means it has Delta X V, not SVT. But it's missing the acceleration. So notice that there's a pattern here. Every single one of these equations is missing. One of the variables except the only one that I have it all in common is the initial velocity. So, guys, what you need to know about these five variables here is that to solve any motion problem, whether it's a car or rocket or whatever it is with these equations, you're always going to need three out of the five variables to solve. So the whole game here. All these problems is figuring out which of these five variables you have in which three you can figure out in order to pick the right equation to get whatever you're missing. So in order to show you how that works, let's just go through this example. Together we have a racing car that is starting from rests. It accelerates constantly, which means that we're able to use the, um equations, which is good. We have 160 m track and the car crosses after eight seconds. We're gonna figure out what's the acceleration of the car. So all of these problems, we can always follow these list of steps to get the right answer. Basically, we're gonna start off by drawing the diagram and listing off our variables. It's a great visual way to figure out what we have and what we need. So you've got the initial velocity you're saying starts from rest. We got this car here, it's gonna travel, and it's gonna cross the finish line, and then it's gonna be moving. Uh, probably, you know, with some final velocity over here. So let's list off our variables. So start off with Delta X then we've got V knots. We've got V A and T now, basically, now that we've drawn and listed the five variables, let's just identify what we know and what our target is. We're told that the track is 160 m, so that's the length that's 160. The time is gonna be, uh, crosses the finish line after eight seconds. So that's the time that's t equals eight. And so what are we trying to find? We're trying to find the acceleration. So this is gonna be our target, Variable A. That's gonna be the question, mark. So that's what we've got here. But notice how we Onley end up with two out of the five variables. Now, whenever this happens, there's gonna be a clue inside of the problem that's going to give you that third one. So let's look at here. We have a racing car that starts from rest and accelerates constantly. So what you need to know there is from rest means that the initial velocity is equal to zero. It starts off with a V not of zero. So that's that third variable that you need so RV not is equal to zero, and now we have three out of five variables. So we're good to go. We can pick an equation. So now the next step here is we have to pick an equation without this ignored variable. So let's take a look what happened here. I was given three variables. Whether numbers or words in the problem, I'm asked for one of them, but this final velocity I have no information about its not asked, and it's not given. So this is called the ignored variable. Over here, it's a variable that has not asked for or given. And so whenever you are picking the equation, you have to pick the equation that contains the variable you're looking for. So, for instance, I'm gonna pick an equation that contains acceleration. But it excludes the ignored variable here. So basically, I'm gonna pick an equation that does not have V inside of it. So let's take a look at our list and figure out which equation that is. Well, the first one has V, so it's not gonna be that one second one says v squared. So it's not gonna be that one. The third one has a But it does not include V, which is good and just to be, you know, thorough. The fourth equation has v plus v not over to. So that's also bad. So notice how this always will give you just one equation to to use. So from this list here, we're gonna pick equation number three. Delta X is Vienna, T plus one half A T squared. So now let's just go ahead and fill out all the variables. So all the numbers we got 1 60 initial velocity zero times eight. So that just goes away. Plus one half. Now we have a you know, we have eight square notice How we're Onley ended up with one unknown variable. That's the acceleration. That's what we want. And then we just rearranging solve. So this one half goes to the other side. Eight square goes to the other side. So I have 160 times two divided by eight squared equals the acceleration. You go ahead and plug that in. What you're gonna get is 5 m per second squared. So that's our answer over here. Moving on to part B. So actually it was part a so part B. Now we're going to be looking for what? The cars velocity is at the finish line. So in other words, we're gonna be looking for what the final velocity is. This is actually the ignored variable in part A. So we're looking for the final velocity over here. You can go through the list of steps. We already have the diagram. We have the five variables. We actually know what this acceleration is Now. Now it's just equal to five target variables. V. So actually, there is no ignored variable anymore because we have four out of the five. That's what makes these problems easier is that as you continue to solve, you figure out more and more the variables. So really, we could start off with anyone of the equations, and the easiest one is gonna be the first one. So I'm looking for the final velocity. So I need V notes, plus a times t. I know V not. That's just equal to zero. I have the acceleration from the first part, and I have time. So I'm just gonna plug this stuff in really straightforward. I've got V equals zero plus and then I've got five from my acceleration and then eight seconds. So we end up with a final velocity of 40 m per second. Alright, guys. So that's it. We're gonna get a lot more practice with this. That's it for this one. Let me know if you have any questions.

2

Problem

A moving hockey puck encounters a patch of rough ice and slides 90cm before coming to a full stop. Assuming a constant deceleration of 8 m/s^{2}, what was the puck’s initial speed?

A

3.79 m/s

B

2.68 m/s

C

37.9 m/s

D

1.90 m/s

3

Problem

Certain rifles can fire a bullet with a speed of 970 m/s just as it leaves the muzzle (this speed is called the muzzle velocity). The muzzle is 71.0 cm long and the bullet is accelerated uniformly from rest within it.

For how long (in ms) is the bullet in the muzzle?

A

144 ms

B

1.44 ms

C

0.722 ms

D

1440 ms

4

concept

## Acceleration Signs

9m

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Hey, guys. So in your motion problems, you run across this phrase speeding up or going faster, and you might take that to mean that your acceleration is positive. But be careful, because this is not always mean this. Students get this tripped up all the time. So in this video, I'm gonna show you how the acceleration, sign effects, velocity and speed. I'm gonna show you really, really simple way to think about this. Let's check it out. So guys, again positive acceleration does not mean speeding up or going faster. Positive. Accelerating positive acceleration literally just means that your velocity is becoming mawr positive. On the other hand, negative acceleration just means that your velocity is becoming mawr negative. We could just see this from the example. I'm sorry for the equation. We know the acceleration is delta view over Delta T. So positive acceleration means that the change in velocity is positive, a k a. Your velocity becoming more positive. And the negative acceleration means that your velocity is becoming more negative. That just comes from the equation. Let me just show you a bunch of examples to see how this works and where the confusion comes from, so you don't make the same mistakes yourself. So we have all these problems here, we're gonna indicate whether the acceleration is positive or negative. And if speed is increasing or decreasing, we're gonna assume Delta T is for So let's get Let's just get to it. We've got V not equals 10 v Final equals 30. So I'm just drawing a quick little sketch. Here's my V Knots, which is 10. Then at some later time from here to here. I know that my final velocity is gonna be a bigger vector, and this is gonna be 30. So to calculate the acceleration, I need Delta be over Delta T. So my change of velocity is 30 minus 10. That's my initial divided by Delta T, which is four. So this is 20/4, and I get 5 m per second squared. So notice how this is a positive number. That's my acceleration. So that's positive. What about the speed? Well, whenever you think about the speed, it just think of the number of the velocity, the magnitude of velocity, not the actual sign. So I'm going from 10. That's the number 2 30 which is a bigger number again. Forget about the sign, which means that my speed is increasing. Let's take a look at part B. Now I have a velocity that's negative. And a final velocity. That's negative. So I'm actually gonna start going to the left. So this is my initial velocity here. Ivy not is negative. 10. And then my final velocity over here is gonna be negative. 30. I know that this is this Takes some time, which is Delta t. So to calculate my acceleration I need Delta V over Delta T My delta V is gonna be negative. 30. That's my final minus my initial negative. 10. Be careful with signs on Delta T is for so this actually negative. 20/4. So this is actually negative. 5 m per second squared. So my accelerations Negative. So this is my acceleration over here. What about the speed? Well, again, Just think about the number, not the sign. Just think about the magnitude of your velocity. I'm going from negative 10. So the number is 10 to negative 30. The number is 30. The number gets bigger. Forget about the sign. So that means my speed is increasing. So notice how one situation. My acceleration was positive. My speed was increasing, and in the other situation, my my acceleration was negative, but I was still speeding up and my speed was still increasing. So speeding up just means that you're going faster. And that means that the magnitude of your velocity just think of the number is increasing. And so the easiest way to think about this is if you're acceleration and your initial velocity are in the same sign, have the same sign. That's when your speed increases. For example, here my acceleration was positive. My initial velocity was positive, so my speed increased over here. My acceleration was negative. My initial velocity was negative. And so my speed increases. Well, let's take a look. More more problems. So here we've gotten a velocity of 30 and a final velocity of 10. So here, now I've got my initial velocity is 30 and then my final velocity is gonna be 10 over here. So I'm gonna drop my little interval from here to here. Now I just calculate the acceleration. A equals Delta V over Delta t my Delta V. Now my final velocities. 10. My initial velocity is 30 and my time is four. So this is gonna be negative. 20. And this is gonna be four. So I get an acceleration of negative 5 m per second squared. So my acceleration is negative. What about the speed again? Think about the number first. I'm going from 30 and now I'm going to tend. The number is getting smaller this time, so my speed is actually decreasing. Let's go. The last one negative. 32. Negative. 10. So here what happens is first I'm going. This is my vision on its negative. 30. Draw it again. Negative. 30 over here. And then finally it's gonna be negative. 10. And then my interval is over here. I know. That's four seconds. So what's my acceleration? So acceleration. Delta V over Delta t. So my Delta V, what's my final velocity? It's negative. 10. My initial velocity. Negative. 30 again. Keep track of all the signs over four. So this becomes positive. Positive. This becomes 20/4 and that's just 5 m per second squared. It's positive. So here we got a positive acceleration. What about the speed? Speed is going from 30 to 10. The number just forget about the sign. The number is going from 30 to 10 which means that my speed is actually decreasing here. So again now we have a situation where we have negative acceleration and decreasing speed. But positive acceleration could also mean decreasing speed So slow slowing down just means that obviously you're going slower. Which means that the magnitude of your velocity is decreasing and this happens whenever you're acceleration and your initial velocity actually have the opposite signs. So for example, here my acceleration was negative. Positive initial velocity. So I slowed down here. My acceleration was positive. Initial velocity was negative. And so that means I'm also slowing down. Alright, guys, hopefully it's sort of like illustrates where that confusion comes from. And so hopefully these these rules here give you a better understanding of what the relationship between signs of acceleration and velocity are. Alright, I got this one last problem here. Let's just get to it. We've got the driver of a truck that is moving to the left of 30 m per second, slows down by taking their foot off the pedal and basically we're gonna calculate the magnitude and direction of the acceleration and it's assumed constant. So we're actually just gonna use our normal steps for solving motion problems with acceleration. So let's just get to it. Let's just draw a quick little sketch here. Eso I've got this truck that's moving to the left. So here I got this truck that's moving to the left. We know that this initial velocity here is gonna be 30 m per second, but if it's moving to the left, it's actually going to be a negative. And so we're told that the truck comes to a stop after traveling 150 m. So basically, after some distance over here, which we know Delta X is gonna be negative 150. Now, at this point here, the final velocity is going to be zero. But again, this is negative because we're moving to the left. So we need three out of five variables. Let's just go ahead and list them out. I need Delta X. I need V not V a n t. I know. Delta X is negative. 1 50. I know my initial velocity is negative. 30. My final velocity is zero. And then my acceleration is what I'm looking for this is my target variable. And then, um yeah, so I've got my diagram, and I've got my five variables. I've got the known variables and what my target variables are. So now I just have to pick the, um, equation that does not have my ignored one. So these are my three variables that I know I'm looking for the acceleration and the time is gonna be my ignored variable. So if you look through our list of equations here, the one that doesn't include time is actually gonna be the second one over here. So we're gonna use equation number two. Whoops. So the the final squared equals the initial squared plus to a times Delta X. And so we're looking for this acceleration. Over here, we know that my final velocity is going to be +00 squared. My initial velocity is gonna be negative. 30 but it's squared and then plus two eight times Delta X. So I get zero equals this into being 900 plus two times a times my delta X, which is negative 150. So this ends up being when I move this to the other side to get negative. 900 equals on. Then what? I multiply too. And the negative 1 50. This becomes negative 300 times a and so all I have to do now is divide and basically my A is going to be 900 over 300 which is just 3 m per second. What about the sign? Well, now my sign is positive. It's a positive 3 m per second, but notice how everything in my diagram here was negative. So even though I was slowing down by taking my foot off the pedal, my acceleration was actually positive. And that's because everything in my diagram was pointing to the left. Alright, guys, that's it for this one. Hopefully, this sort of clears up some of the confusion. Let me know if you have any questions.

5

concept

## Acceleration with Multiple Parts

8m

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Hey, guys, For this video, we're gonna check out how we solve problems where an object is moving under constant acceleration in one or multiple parts. And what we're gonna find is that it's exactly like how we solve any motion problem in general. So let's check it out. Now, remember, in these problems, organization is our friend here, so the situation is gonna be different. You got a guy running and speeding up or slowing down. But when there's multiple parts, the first thing you want to do is just draw the diagram and then list of five variables like any motion problem. So you have parts A to B and then B to C, and then we also have the whole entire interval over here, and now we just have to draw. Now we have to list are five variables. So we've got Delta X initial velocity, final velocity acceleration in time, But because there's gonna be two sets of them, I'm gonna have to assign them variables. So I'm just gonna give them little A to be. Now, my initial velocity is actually gonna be the velocity right here. So I'm gonna call this V a final velocity is gonna be right here. So I'm gonna call this V B. And then you've got the acceleration in the time. Now we just do the exact same thing for the second set. So Delta X from B to C. Now, my initial velocity is VB. My final velocity is V c and then the acceleration and time. And now, finally, you also remember we have the two different parts. But we also have the whole the whole entire interval and the only two variables that matter. There are Delta X from A to C and then delta T from A to C. So it's the total distance in the total time. Usually those are the two ones that are given or there I asked for. So once we have this or sort of organizational structure for all of our problems and we have the list of variables now the rest of us just picking equations. So let's just work out this example together and see how this stuff works. So we're told that we're driving at a constant 30 m per second and then we see a road hazard and it takes 300.7 seconds for us to react and stop. But then we decelerate and then come to a stop. So that means that there are two different intervals going on here. So the first step is always to set up the problem by drawing the diagram and then listing your five variables. So let's go ahead and do that. So I've got this little diagram here. This is the part where I haven't yet slammed the brakes and it takes 0.7 seconds for me to react. So this is the reacting part. And then also over here I've got Once I've slammed the brakes, Now I'm coming to a stop. So this is the stopping part. So now we're just gonna go ahead and list the five variables Delta X from A to B once we've actually labeled them right. This is a B and C, and now I've got my V A V b acceleration and time. Okay, so I don't know the distance for this first part. All I know is that I'm driving in a constant 30 m per second, which is my initial velocity. So this is right here and because it is a constant 30 m per second, that's also gonna be my final velocity when I get to the end of this first interval, which means that the acceleration is equal to zero. So acceleration of A to B is zero the same exact velocity. And we also know that it takes 00.7 seconds first to react and stop. So those are our numbers there. Now, let's do the second, uh, the same thing for the second interval. I don't know the distance from B to C. I don't know the stopping distance. All I know is that the initial velocity here is gonna be 30 m per second because that's what it is over here. And then I know that I'm gonna come to a stop. So a stop means that the final velocity is gonna be zero. And I know that the acceleration in this point is gonna be 6 m per second. But it's decelerating, so it's gonna be negative six and then finally, the time. I don't know how long it takes for us to come to a stop. So these are all of our basically are variables here. So the second step is for each part, we just have to identify the known and now with the target variables. So what is part a asking us for? It's asking us for how far we're traveling before we're applying the brakes. So which one of these 10 variables is that gonna be? Well, before the brakes is actually in this first part over here, and it's asking us for a distance. So it's actually asking us for Delta X A to B. So that's that second step there so we know what we're looking for. Delta X from A to B. Now the third step is just to pick the right equation that does not have are ignored variable. But it's a little tricky. Here is a little weird here because we actually have four of the five variables, and more importantly, the acceleration is equal to zero. And so, if you go to our list of equations here, remember that we have two kinds of motion equations. When the acceleration is equal to zero, there's Onley just one. It's constant velocity, so we actually don't have to pick any equations here because there's really only just one to use. So that's V. Abe is Delta X from A to B over ta to be. And if we can actually re arrange for this and then sulfur Delta X So this is gonna be V A B times ta be. So this is just gonna be 30 times 300.7. And if you work this out, this is gonna be 21 m. So that's the answer. So it's just 21 m and that's the distance. So let's move on to part B now and go through the list of steps. We've already drawn the diagram, and now we just have to figure out what the target variable is. So now it's asking us in part B for the time it takes to stop after applying the brakes. So which one of the 10 variables that apply to Well, it's a time that's after we apply the brakes. So it's gonna be in the second interval, So that's actually gonna be the time. From B to C. This is gonna be the variable that we're looking for here. So that's the second step. The third step now is picking the, um, equation without the ignored variable. So if you take a look here, we have to use our, um equations because the acceleration is not equal to zero. But if you'll notice Also, we we have three out of the five variables and this is the variable that we're looking for. So that means that they ignored one is gonna be the Delta X from B to C. I have nothing no information about it. And so the equation that does not have Delta X is gonna be the first equation. So that's the equation I'm gonna use. So one this the final velocity, which is V C. Is equal to the initial velocity V B plus the acceleration times time. So if you go through our list of variables here, I know V. C is zero. I know VB and I have the acceleration. The only unknown here is time, so I've got zero equals 30 plus negative six times TBC If you go ahead and work this out really quickly Uh, the time from B to C that's equal to five seconds. All right, so that's our answer. So now we know that this is actually equal to five seconds. So now finally, let's move on to part C. So again, let go through a list of steps. What is the target variable in part C. It's actually asking us for the total distance traveled. So remember the total distance is going to be total distance for the entire interval from A to C. That's gonna be Delta X from A to C. And remember that Delta X from A to C is really just gonna be a B and B c added together. Right? Remember, if this is 10 and this is 20 then the whole thing is 30. That's just an example. So basically an equation where you can use is Delta X from A to C is equal to Delta X from a B plus Delta X B to C. Some of the words it's just 21 because already know this plus whatever b to C is and this is gonna be my final answer. So I'm just gonna go real quick and figure out what that Delta X from B to C is. That's all I have to do. So let's go over here and do that. Delta X from B to C is what? Well, so now this is my target variable. So now I just have to pick the equation that does not have the ignored variable. But now, if you've taken a look here, I already have four out of the five variables, so there is no longer and ignored variable. And whenever this happens, you can actually pick any of the equations that has Delta X in it, right? It's the Onley, other unknown that's left so you can pick any one of these equations here that has Delta X. The choice is really up to you and you'll get the same answer every single time. So what I'm gonna use is I'm gonna use equation number three so Delta X b c is gonna be the initial velocity V b times TBC plus one half a times t squared and I actually have everything right. I have four of the five variables, so I'm just gonna plug it in. So Delta X from B to C is gonna be 30 times five plus one half times negative six times five squared. And if you work this out in the calculator, what you're gonna get, you're gonna get 75 m. So we just plugged that right back into their for the total distance. And so we have 21 plus 75 and that's going to equal 96 m and that is our final answer. Alright, guys, let me know if you have any questions, let's get some more practice.

6

example

## Subway Train

10m

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Hey, guys, let's check out this problem together. We've got a subway train and it's gonna start from rest and accelerates. Then it's gonna travel a constant speed for some time. And then finally, it's gonna slow down until it stops to the next station. So there's a bunch of moving parts here with multiple accelerations. So we're gonna solve this like any of the problem with multiple parts with accelerations. First, when you have to do first 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, ah, called Adobe. Then it's gonna travel a constant speed. So from B to C, it's gonna be constant speed. So this is constant. And then finally, it's going to stop I'll as it approaches the next station. So this is actually gonna be from C to D, and it's gonna 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 Delta X from A to B, which we know which we actually don't know we don't know Delta X me to be That's the distance In the first part we've got the velocity A and Velocity final So we know that this thing is gonna start from rest So this is just gonna be zero and let's see s We know this is zero We don't know the final velocity but we do know the acceleration of 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 gonna 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 gonna accelerates, which means it's gonna speed up in the right direction. So that's gonna be positive. This is gonna be 1.5 positive. Okay, for the second part now we've got constant speed, so constant speed for third for 60 seconds. So we've got Delta X from B to C. We don't know what that is. VB. We don't know what that easy either, because we don't know what it is from the first part v c don't know what that is. Acceleration from B to C is gonna be well, it says a constant speed. So this constant speed here actually means the acceleration is equal to zero for this part. But we do also know that the time for part B, it's 60 seconds. So whatever this velocity ends up being VB in B. C, they're actually gonna be the same thing because there's no acceleration. All right, so finally, now we have this stopping part. So we have Delta X from CDD. We don't know what that is. V c don't know what that is. V d. That's just gonna be when it stops the next station right here. So this is gonna be zero. And then finally, the acceleration from C to D Well, it's going to slow down at 3.5 m per second squared, so first accelerated. Then it went to constant speed. And then it's going to slow down, which means that this acceleration is gonna 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 what's the total distance that the train covers. So I remember we have all of the different parts of this of this motion here, but we also have the whole entire interval and this whole entire interval here there is a distance. Delta X 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 Delta X from a D. D. Is our target variable. How do we figure that out? Well, we don't have an equation for Delta X, A two D using our motion equations, but we can't figure it out by piecing together all of the different distances in each of the parts. So right, if this was 10 2030 you would just add them all together. That would be your total distance. So this guy is actually gonna be Delta X from A to B plus Delta X from B to C plus Delta X from C to D. So we're gonna 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 gonna look for Delta X from A to B. So this is gonna 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, um, equation, one of our motion equations that doesn't have the ignored variable notice how I have three out of five variables 123 and which means the one I'm ignoring is actually gonna 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 gonna be well, let's see, we have VB here, this final velocity Final velocity and this is also gonna be final velocity. So we're actually gonna pick equation number three So this is going to say that my Delta X from A to B which is actually my target variable, which is great because this is gonna be on the left side of the equation is gonna be my initial velocity times time. Well, we know this is gonna be zero, because our va is equal to zero plus one half acceleration, which is 1.5 times t squared. So this is gonna be my 14 seconds squared, so I could just actually plugged that straight into my calculator on my delta X from A to B is gonna be 147 m. So that's one of my variables. 1 47. 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 gonna be my target variable here, Delta X from B to C So we'll go and focus on that. So how do I figure this out? Well, my Delta X from B to C So which equation is not gonna use will remember that we only use the four, um, equations whenever the acceleration is constant. And it's not zero when the acceleration is equal to zero that all of our equation simplify. And we only just use one, which is V equals Delta X over Delta t So we're trying to figure out by Delta X from B u c We're just gonna use the constant velocity formulas. My V B to C is equal to Delta X B two C over T B to C, which means that Delta X from B to C is just gonna 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's all I have to do is just figure out what the velocity is from B to C and notice how it's actually gonna be the same velocity. Um, it's gonna be either VB or B C because it's gonna be a constant velocity, right? So whatever I figured out for VB is gonna be the same thing for for B. C. 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 gonna have to go to a different interval and figure that out. Which other interval also has VB? Well, these either. Actually, there's only one choice. It's gonna be the first part right here. So what we could do is we can actually use the first interval because now we know four of the five variables to figure out what my velocity is that I ignored in the first part. So that's what we're gonna go ahead and do. So now what I can do is I could go back over here, and I could say, Well, now I want to figure out what my velocity B is. So how do I do that? Well, now, I'm gonna use my three out of five variables for this part here. And if I'm trying to figure the final velocity, I could just use the first equation. That's the easiest one. So I'm gonna use equation number one to figure out the final velocity for VB is equal to V A plus a times t. So I know I'm going from rest. So this is zero, and then my acceleration is 1.5 and my time is 14. So I get 21 m per second. So that means that I know that this is 21 this is also 21 which means now I could go ahead and use my formulas. Now I have the velocity and I have the time, which means I can figure out Delta X. So this history is gonna be 21 times my t B C, which is 60. If you work this out, you're gonna get 12. 60. So notice how sometimes you have to stop and you're gonna 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 you think about, like, pieces of a puzzle. I've got the second piece of the puzzle 12. 60. So now I just have to figure the last one. The last one is Delta X from C to D, So I gotta figure out this guy over here Notice how also, I figured out another variable when I did this. So remember how V c was an unknown. Well, we just figure out that V. C was 21 here, traveled a 21 m per second. So that means it ended right here and now it's 21 m per second over here and then eventually is going to stop. So now we're looking for the Delta X from C to D. We just go ahead and look at our, um, equations. We know there is an acceleration here, so we are going to use these variables. Which means we do need three out of five. I've got three out of 5. 21 0 and negative 3.5, which means that my t is gonna be my ignored variable over here. And so now, which equation we're gonna use? Well, the one that t is ignored in is gonna be equation number two. So this is when we're using for two for for a Delta X from C. D. So this is just saying that my final velocity V D squared is equal to my initial velocity V C squared, plus two times a from C d d times Delta X from C to D. So this is actually what I'm looking for here My Delta X. So I just work everything else out, you plug everything else in. So we know my V d is gonna be zero. My V c is gonna be 21 and then two times the acceleration. Well, the acceleration is gonna be negative. 3.5. Remember the negative sign and then times Delta X from CDD. 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 gonna get negative. 4. 41 equals and then you multiply these things out. You're gonna get negative seven times Delta X CD. So now my Delta X M. C. Diddy is just Well, the negatives will cancel. You're just gonna get 4. 41/7, which is equal to 63 m. And so this is the final piece of the puzzle. The final displacement. So now you just go ahead here, you put them all back together again. So 63 I can say that Delta X from a D. D. The total distance traveled of all of the parts is gonna be 1 47 plus 12 plus 63. You add all these things together and What you get is you get 14 70 m 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 gonna get totally lost when you're trying to figure this out. Anyway, let me know if you have any questions. That's it this way.

Additional resources for Kinematics Equations

PRACTICE PROBLEMS AND ACTIVITIES (16)

- A speed skater moving to the left across frictionless ice at 8.0 m/s hits a 5.0-m-wide patch of rough ice. She...
- a. Find an expression for the minimum stopping distance dₛₜₒₚ of a car traveling at speed v₀ if the driver’s r...
- A car starts from rest at a stop sign. It accelerates at 4.0 m/s² for 6.0 s, coasts for 2.0 s, and then slows ...
- You’re driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you....
- A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After travel...
- A sprinter can accelerate with constant acceleration for 4.0 s before reaching top speed. He can run the 100 m...
- A motorist is driving at 20 m/s when she sees that a traffic light 200 m ahead has just turned red. She knows ...
- At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161...
- An astronaut has left the International Space Station to test a new space scooter. Her partner measures the fo...
- A pilot who accelerates at more than 4g begins to “gray out” but doesn’t completely lose consciousness. (b) Ho...
- A pilot who accelerates at more than 4g begins to “gray out” but doesn’t completely lose consciousness. (a) As...
- In the fastest measured tennis serve, the ball left the racquet at 73.14 m/s. A served tennis ball is typicall...
- A Fast Pitch. The fastest measured pitched baseball left the pitcher’s hand at a speed of 45.0 m/s. If the pit...
- An antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 6.00 s. I...
- The human body can survive an acceleration trauma incident (sudden stop) if the magnitude of the acceleration ...
- For an object starting from rest and accelerating with constant acceleration, distance traveled is proportiona...