2. 1D Motion / Kinematics
Intro to Velocity and Speed
Was this helpful?
Hey guys in previous videos, we talked about the difference between distance and displacement to describe how far something moves. And in this video we're gonna talk about another pair of important but related words called velocity and speed. Let's check it out. So let's take a look at a familiar example, let's say we were moving from A to B and then back to C. Again, we know that this is 10 m to the right and we're going to move six m and then we're gonna be moving to the left here, there was two words to describe how far something moves. That was distance, which remember was given by the letter D. And was a scalar your distance just with the total of everything that you travel 10 and six. So we know this is just 10 and six and that makes 16 The displacement on the other hand, was a vector and it's the difference between where you started and where you ended and it was the shortest path between those two points. And we know that your final position is for your initial position is zero. So your displacement, which is just 4 0, was just 4m to the right. So just like distance versus displacement, there's two words to describe how fast something is moving. And the basic idea here guys, the main idea is that speed is related to the distance that you travel, whereas velocity is related to the displacement that's traveled. Let's take a look. So for the equations, speed is going to be s and it's defined as the distance over time. Whereas velocity is defined by the letter V. And it's defined as the displacement over time. So for the equations, the speed is gonna be distance over time, D over delta T. Whereas velocity is gonna be delta X. The displacement over delta T. And they're both given in terms of units of meters per second, that's gonna be the units that were working with over here. So let's take a look. So we've got that this distance here, the distance remember is going to be a scalar which means that the speed is also going to be a scalar. If this is a scalar then that means that this is going to be a scalar. Now on the other hand, the displacement was a vector. And if the velocity displacement over time and that means that velocity is also going to be a vector. So scalar for speed and vector for velocity. Now let's talk about the signs now, for speed, the speed is always going to be either positive or zero because it depends on the distance and the distance can always just be positive or zero, it can never be negative, whereas the displacement can be positive negative or zero. So your velocity can be positive negative Or zero because that's what your displacement can be. And so let's talk about negative velocity for a second, what does negative velocity mean? Remember signs in physics just have to do with directions. So negative just means that you're moving in the opposite to whatever the positive direction is and your problem is gonna tell you what that direction is. So for example, let's say the right direction is positive, anything moving to the left is going to be negative, and sometimes in rare occasions you might see that the left direction is positive. Your problem will tell you this, in which case anything that's moving to the right is going to be negative. So just means whatever your problem tells you the positive direction is anything moving to the opposite of that is gonna be negative velocity. Alright guys, that's really all there is to it. So let's check, let's take a look at an example, we're gonna jog 15 m in two seconds, then nine m backwards in another two seconds. Let's calculate the speed and velocity. So let's just draw a quick little diagram of what's going on here. So I'm gonna move, I know this is going to be 15 m in this direction, and then I'm gonna move 9m backwards in another two seconds. So this is gonna be 9m like this. So I know that the time here is going to be two seconds and the time here is also gonna be another two seconds. So let's say this part, a the speed is going to be, remember the formula we're gonna use is the distance over delta T. That's the equation that we have right up there. So we're gonna use distance over time. Now, we just have to figure out the distance for the total trip. So what's the total distance? Will we move 15 m forward and then 19 m backwards? So your total distance is gonna be 15 plus nine m and your total time, is that two seconds or no? Because remember this is two seconds for the forward direction and in two seconds for the backwards direction. So our total time is two plus two. And so what we get is 24 divided by four. So that means our speed was six m per second. What about our velocity velocity member is delta X over delta T. For the whole trip. So now what happens is we need to figure out what's the shortest path from initial to final. So just in the same way that we did the example above this is A. And B. And this is see the shortest path between initial and final is actually gonna be from here to here. So this is my displacement, this is gonna be delta X. And what is delta X. Delta X is 15 m forwards and nine m backwards. So we can do 15 minus nine over here. And then that just gives us six m. So our displacement is actually 15 minus 9/ plus two, which is the total time. So we're gonna get 6/2. And what we actually get is uh whoops, that's equals, I'm sorry, 6/4, whoops. And so we get is 1.5 m/s. Now, we have to specify a direction because we've got a positive number here. That makes sense because our displacement also points to the right, so that means our velocity vector also points to the right. That's it for this one, guys, this is the speed and velocity. And let me know if you guys have any questions.
Beginning from a signpost, you run 60m to the right, then 60m back. The entire trip takes 24 seconds. What is your speed and velocity for the whole trip?
s=5 m/s, v=5 m/s
s=2.5 m/s, v=0 m/s
s=5 m/s, v=0 m/s
s=0 m/s, v=0 m/s
Solving Constant and Average Velocity Problems
Was this helpful?
Hey, guys. So oftentimes, in physics, you have to solve problems in which there is a constant or average velocity. So in this video, I'll show you how to solve these problems by using a slightly different form of the velocity equation which are already pretty familiar with. Let's check it out. Do you guys remember? The velocity is measured between two points an initial to a final position. So your textbooks might prefer this as an average velocity. Let's check it out. If you were going from 0 to 8, for example, in a span of four seconds, then your velocity is just Delta X, which is a change of eight over Delta T, which is a change in four. And so you get 2 m per second. But this velocity is often called an average velocity because you don't really know what happens in between. All you know is that you went from 0 to 8 and it took you four seconds. But during the middle, you could have sped up to 3 m per second, slowed down toe one and then sped back up again to three. You have no idea if you were to plot this out on a velocity versus time diagram that shows your velocity and y axis and your time in the X axis. It would look all squiggly like this because you're speeding up and slowing down. And this makes physics problems a nightmare to solve, because you have to sort of account for all the speeding up and slowing down. So one way we can sort of go around this is we can say, Well, we can kind of smooth out all of these velocities and pretend as if we were just traveling with a constant or average velocity that Smoothes out all of these different variations. So this is an average velocity. This makes things a lot easier in physics, because average velocities behave like a constant velocity in constant velocity. Problems in physics mean that there is no acceleration. And whenever there's no acceleration, there's only one equation that we're gonna use. And it's the one we already know. It's that V average or the velocity is just equal to Delta X over Delta T. That's it. That's the one equation that we're gonna use. Now we can take this equation and we can actually manipulate it to solve for Delta X because there's some problems. We have to be working with Delta X in the position. And so if we re arrange for this, we've got Delta X equals V average times Delta T All they did was just just move the delta t to the other side. And then some problems. There are some textbooks will start dropping this average over here. Now we can also do is take this Delta X and we know that this is just the X final minus X initial. So some textbooks and some equations will move this over to the other side. And what we get is we get this equation over here. X equals X not plus v times Delta t So again, this really is just a different form of this equation kind of just expanded out, and it gets a special name because it gets used a lot in physics. It's called the position equation. So later on, we're talking about the position equation. I'm talking about this guy over here, and it's called the position equation because it tells us that your final position X is just your initial position. X not plus how far you've moved that v times Delta T. And so we're gonna use this equation over here whenever we're solving for the velocity and we'll use this equation here. Whenever we're trying to solve the positions, that's all there is to it. Let's get some practice, guys. All right, so we're gonna solve for the unknown variables in each one of these diagrams here we've got from going from 0 to 20 and five seconds and we want to calculate velocity. So that means that we're just gonna go ahead and use the velocity equation here. So that means that our velocity is gonna be Delta X over Delta T. And so Delta X is 20 minus zero, right? That's my initial and final divided by a change in time of five seconds. So I just get 4 m per second. That's pretty straightforward. Let's move on. So now I've got my initial position, velocity and time, and I'm trying to find my final position. So now I'm gonna use my position equation just because it already gives me what the final position is gonna be is variable. So I've got X is gonna equal X not plus v times delta T So my final position is my initial position of two plus the velocity of three. And the change in time of six seconds. Three times six is 18 to plus 18 equals 20 m. So this is gonna be my final position right over here and for the last one. Let's check this out. Now we have to be a little bit careful, because if you notice in the first two diagrams who are moving to the right But in this last one here, we're actually moving to the left. So we have to be careful with that because we know this is gonna be the negative direction. This is gonna be the positive direction, and that makes sense because their velocity ends up being negative here. But we're still just looking for the initial position now. So we're gonna start off by using the position equation. X equals X not plus V times Delta T. So if I'm looking for this initial position, then I could just move all of this over to the other side. So I get X minus V times Delta T is equal to X. Not so. Let's just make sure I have everything I know what my final position is. I know what the velocity is. What about the change in time? Delta T? Well, I have my final and initial times. That's not Delta T. So I could just quickly go ahead and say, Well, these two Delta teas, they're gonna be t final minus t initial. So this is just gonna be seven minus three, which is four seconds. So now I have my delta t and now I can go ahead and plug everything in, so my final position is gonna be negative. 20. You plug it in with the signs and be careful minus my initial of my Sorry, My velocity is negative. Four again. Be careful with the signs. And then delta T is four seconds. And so these end up canceling out so you can You can make them both positive and you get negative. 20 plus 16 is equal to Delta are my initial position. And if you work this out, this ends up being negative 4 m. And that's our answer. That's our initial position. Alright, guys, that's it for this one. Let's 4 m. Let me know if you guys have any questions
Was this helpful?
Hey, guys, let's check this problem out. We have a baseball pitcher who can throw a baseball at some speed. We're trying to figure out how long it takes for the baseball to travel the distance to the home plate. So let's just draw a quick little sketch. I've got a baseball like this, and I know that the velocity that this is thrown out is 44 m per second. Now. I'm not told that the baseball slows down, speed up or anything. So when this happens in problems, you can assume that this is a constant velocity. So it's a constant velocity that the baseball is traveling, and we know that the distance to the home plate, which is Delta X, is gonna be 18.5 m. So now the only thing we have to figure out is what's the time? So what is the time while we're working with constant velocity, which means we could only use one equation now, that equation, remember, is that the whoops that's the velocity is equal to Delta X over Delta T. And so what's the variable that I'm looking for? It is just delta T, so I could just re arrange for this really quickly so I could move this. These basically to the other side. I could swap the V and the Delta T. And what I get is the T is equal to Delta X, divided by the velocity. So that's just 18.5 divided by 44. And if you work that on your calculator, you're gonna get 0.42 seconds. So about half a second for the baseball to travel to the home plate. So for those of you baseball players out there, uh, this actually might make some sense to you. Alright, guys, that's it for this one. Let me know if you have any questions.
Constant Velocity with Multiple Parts
Was this helpful?
Hey, guys. So a common type of problem that you'll see in motion involves an object that is moving with different but constant velocities in multiple parts. For example, let's say we had a guy that was walking from a to B with a constant velocity, then picks up speed with a different but constant velocity from B to see what makes these problems hard is not the math, because at the end of the day, there's no acceleration, these problems, So we're only just gonna be using this this equation here and three variables. What makes these problems heart is actually organizing which variables you have and which ones you need. So I'm gonna show you a list of steps out down here that we're gonna use to solve any one of these problems. Let's check it out. And the key to solving these problems is again organization. So we're gonna list off the three variables after we draw the diagram. We're gonna do this every single time. We've got three variables to keep track of V, Delta X and Delta T the Delta X and Delta T. But because there's two sets of them, what I'm gonna do is I'm gonna attach letters to every single one of them. So this is gonna be V from A to B Delta X from A to B and T from A to B do this every single time v from B to C Delta X from B to C and T from BBC. Now you'll never lose track of which variables you have in which ones you need. There's also before we move on, actually a third interval that we have to keep track of. There's not just the two pieces. There's also just the whole from start to finish. And there's also equations over here in variables that we need V from A to C Delta X from A to C and Delta T from a d. C. So this is what we're gonna do for all of our problems. Draw the diagram, list out the variables and then figure out what we need to know. Let's go ahead and solve the problem. So we've got a car that's traveling at a constant 50 m per second for 10 seconds, and then it changes and it goes at 30 m per second for 600. So this is constant velocity, multiple parts. So we're gonna list off the intervals A, B and C, and we're gonna draw the diagram and then list out the variables for all the pieces. So you got V from A to B? We've got Delta X from A to B and we've got Delta T from A to B. I know the velocity in this first part here is 50 m and I know the time is equal to 10 seconds, but Delta X I don't know now from B to C do the same thing. The from B to C I know is 30 Delta X from B to C. I know it's 600 m here, but the time is something I don't know. Delta T from B to C is unknown. So let's take a look at this first part here. The first part asked us to find the total distance traveled, which variables that will remember. There's a third interval here. It's the one from start to finish. We need the three variables here V from a D. C. Delta X from A to Z and Delta T from ABC. So this is actually the variable that we're looking for what's the total distance off all the parts combined? And so this is gonna be our target Variable here. Now let's move on to the second step, which is where Right equations for each one of the intervals here. Now there is only just the one constant velocity equations. But there are a couple more equations that weaken right. For instance, this total distance traveled here, if you have multiple parts, is really just the distance that you cover in the first half from A to B plus the distance that you cover in the second half from B to C. For example, if this was 10 and this was 20 then the total distance was 30. That's just an example. So let's see here. My total distance from A to B is actually unknown, so I've got a big question mark your here. But then my total distance or my displacement from B to C, is equal to 600. So if I can figure out that first half, then I can figure out the total over here, and that's what we're gonna go ahead and find out. So I've got to figure out what this Delta X from A to B is. So now I'm just gonna go ahead and write the equations for the interval. From A to B. There's only one equation that we're gonna use. So we've got these three variables here. My my equation is VB equals Delta X A B over Delta ta be. That's just the average velocity formula. Now, what are we looking for? Remember that we came to this interval because we're looking for Delta X from A to B toe, plug it into this equation. So we just have to solve for that Delta X from A to B is equal to V A b times delta t A B. So in other words, my V is 50 and my time is 10 seconds. So this is just a displacement of 500 m. So now I can plug this back into this formula. Over here, I get 500 and 500 plus 600 gives me the total distance traveled off 1100 m. So this is my answer over here. So that's my answer of 1100. So let's move on now to the second part, the average velocity from start to finish which variable is that Well, the average velocity from start to finish is again going to be in the interval from A to C. So just how this was from A to C. This is also from A to C. So these variable here I'm looking for This was the total distance travel that was Delta Extra May to see. This is my average velocity from start to finish. That's gonna be this variable over here. So now I'm just gonna go ahead and write the equations for each one of my intervals. My V from A to C is gonna be Delta X from A to C over Delta T from A to C. And so if you take a look at this equation here, I just figured this part out. Actually, in the first half, this is just the 1100 that I just solved. For Now I just need to figure out my delta t a c, which I don't know. So now I just need a new equation for this delta T A c Well, if the total distance traveled was basically just the some of the distance over here and over here, the total time that it takes for both of these intervals is gonna be the same thing. What's the total time from A to B plus, what's the total time from B to C? So if this was 10 seconds in 20 seconds, then the total you know, the total time taking was 30 as an example. So now let's look here. Because now I actually have tea from A to B. I know that this is just 10 seconds over here, so I've got 10 plus now. I just need the time from B to C, which I don't know. So now I just need to go and get a new variable in order to get my total time, and then I can plug it back into this equation over here and solve for the velocity. So I know that this is gonna be 1100 divided by whatever I get for that variable over here. And this is gonna be my final answer. All right, so, basically, I need to go ahead and get this tea from B to C over here. So I'm gonna write the equation for this interval. My delta X from B to C. I'm actually sorry. Whips my V from B to C is gonna be Delta X B two C over Delta T B to C. And so what I can do is if I'm solving for this guy over here, then I can move this to the other side and then trade places with the BBC. So my delta t from B to C is gonna be Delta X from B to C, divided by V B two C. So, for example, uh Or so, for instance, in this problem here, my Delta X B to C is 600 my V is equal to 30 which means that my time is just 20 seconds. So I know now this is 20 seconds, which means I can plug it into this formula over here. This is 20 seconds, and so now I can add these two things together. So now the total time that it takes for the whole trip is just 30 seconds. And so now that I have both of these numbers here, now I can get my final or average velocity for the whole entire piece. It's 1100 divided by 30 and this is gonna be 36.6. Repeating and this is gonna be meters per second. So this is my final answer. Alright, guys, that's it for this one. Let me know if you have any questions. Like let's get some more practice.
You walk to the right at 3m/s for 8s, then turn around and walk backwards at 2m/s for some unknown time. You end up 16m to the right from where you started. How long did you walk backwards?
Race Against Time
Was this helpful?
Hey, guys, let's work out this problem together. We're told that a runners trying to complete a race, they're running at some speed for some time. Then they realize they have to go faster. And we have to figure out how fast they have to go in the second part of the race so that they finish in some time. So this is a classic motion with multiple parts. So let's just stick the steps, draw the diagram and list out all the variables. We know that from start to finish, the entire race is going to be 100 m in 20 seconds. But there is a point where they realize they have to go faster. So there is a point right here where they're going to change speeds. So remember, we have to label all the intervals. This is gonna be a B and C. And so I'm gonna list out all the variables for this one in red and this one in blue. So list out the variables we know we're gonna have Delta X from A to B. We're gonna have velocity from A to B and T from A to B. Also noticed that in the problem. It says that we're in constant or average velocity for a whole entire time, so there's really only just those three variables to keep track of now. For the second part, we have Delta X from B to C. We have the velocity from whoops. That's the velocity from B to C and then the time from B C. And now finally, we remember we have the whole entire thing. There's two parts. There's also a whole interval that we have to keep track of. That's this guy over here. And we know that this is the total. We know that the total distance for this race is 100 m, and we also know that the time that we're supposed to complete this race is 20 seconds. So we know what both of those values are for Delta X and T. So let's just go through now and identify the other things we know about the problem. In the first part, we're gonna be running for 4 m per second for 14 seconds. So that means that the velocity from A to B is gonna be four. And the time is gonna be 14. But we don't know what the distance is in the second part. We actually don't know anything about this problem. We don't know anything about this interval. We don't know that the distance, the velocity or the time. In fact, the velocity in the second part is actually what we're trying to figure out. What's the average velocity for the rest of the race to complete it in 20 seconds. So this is gonna be our target Variable over here. So now that we've drawn the diagram and we've listed all the variables now we just have to go ahead and solved by writing the equations for each of the intervals. So remember, we're working with constant velocity. There's only one equation in each interval that we consult for. So the velocity from B to C is gonna be Delta X BTC over T B to C. So, basically, all we have to do is just figure out. Remember, Delta X B C is gonna represent the distance left at the end for the end of the race, and the TBC is the time that we need to finish it toe in order complete under seconds. So I just have to go ahead and figure out what each one of these variables are in order to get that that that average velocity that I need. So whenever I'm stuck and I need to figure out what Delta X B C is, um, there's only one other equation that I could go and find. So if I don't have it over here, then I have to relate it to the total the total distance, which is Delta X from A to C. I know that this distance is 100 m, but I also know that this distance is made up of both of these two distances added together. So basically I can set up in equation Delta X from A to B plus Delta X from B to C. And now I just have to figure out if I actually know two of those things. Obviously, I know the total distance is 100 but if you look through this problem, I don't know what either of these things are. And this is really what I'm trying to look for so I can plug it back into this equation. So can I figure out what this delta X from A to B is That's the question. So can I figure out what this thing is? Well, if I go over to this interval over here, I have two out of the three variables that I need so I can actually solve for Delta X from A to B. So we start off the equation V A to B is equal to Delta X from A to B over ta to be. And so I could rearrange for this really quickly on dykan get that Delta X from A to B is equal to V A B t A B, and I actually know what both of these things are. So I could just go ahead and solve. This is gonna be four times 14 which is gonna be 56. So now I can just pull this back into this formula and figure out what that distances. So I know that Delta let me set up this a little bit differently. I have 56 plus Delta X B C is equal to 100 right? The first part from the second part is plus the second part is equal to 100. So that means that Delta X from B to C is 100 minus which is equal to 44 m. So that's 44 m, which I've got right there. So that Z, that's the first variable that I that I need and the second one that I need now is the time. So remember that there's two. There's two variables that I was missing in this formula. I also need to figure out the time Well, if we go, if we go figure out the time over here the time the first part was seconds. But the time that I need to complete the race in is 20. So this is pretty simple, but basically the time that this needs to be is six. Because I have to finish the race in 20 seconds. So if this is 14 and this has to be six, so now I have everything I need. So the V B C is equal to 44 and then divided by six. And if you work this out, you're gonna get 7.3 m per second. So they have to speed up to the speed in order to finish the race in that time. Alright, guys, that's over. This one. Let me know if you have any questions
Additional resources for Average Velocity
PRACTICE PROBLEMS AND ACTIVITIES (3)
- You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 km/h (65 mi/h),...
- A car is stopped at a traffic light. It then travels along a straight road such that its distance from the lig...
- In an experiment, a shearwater (a seabird) was taken from its nest, flown 5150 km away, and released. The bird...