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Average velocity for constant acceleration

Investigate the relationships between velocity, acceleration, and distance in a physics context. Learn how to use formulas to calculate final velocity and total distance traveled given an initial velocity, constant acceleration, and time. Created by Sal Khan.

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  • male robot hal style avatar for user QuantumParticle
    What is an increasing acceleration called (i.e. ∆a/∆t)?
    (19 votes)
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  • female robot grace style avatar for user santiago sarasola
    in my physics book i have another formula of distance:
    it's initial distance + intial velocity x time +1/2 acceleration x time^2
    is this formula other way of writting the formula of sal?
    (20 votes)
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    • leaf green style avatar for user Aditya Vss
      The formula you listed in the first question is a derivation of the sum of areas of the rectangle and triangle. Here S is displacement, u is initial velocity, v is final velocity, A is acceleration and t is time.
      D = ut + (1/2)(v-u)t
      (Multiplying and dividing t)
      D = ut + (1/2)[(v-u)/t]t.t
      [(v-u)/t is Acceleration]
      D = ut + (1/2)A(t^2)
      This can be also written as D = Initial Velocity x Time + 1/2 Acceleration x Time^2
      (3 votes)
  • hopper cool style avatar for user Arnab Chowdhury
    At why did sal use a triangle before the t?
    (4 votes)
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  • purple pi pink style avatar for user sjkang
    At , I'm having trouble understanding why the equation can also be called as the average velocity. I remember from the previous videos that average velocity is the displacement over time. How can (1/2)(change in t)(v(i)+v(f)) be the average velocity?
    (7 votes)
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    • piceratops ultimate style avatar for user Jairaj Basur
      Sal does not include ∆t in the formula for average velocity. Velocity = Displacement/Time. Average Velocity only equals Displacement/Time when the Velocity is constant. When Velocity is not constant, Average Velocity = Initial Velocity + Final Velocity/2( the average of V(i) and V(f) )
      (5 votes)
  • piceratops seedling style avatar for user Vincent Morgan
    Am I correct in assuming the logic still holds by using the area of a trapezoid, instead of triangle plus square?
    1/2 x (a+b) x h
    So that'd be:
    1/2 x (vi+vf) x t, which is what you got in the video.
    (4 votes)
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  • orange juice squid orange style avatar for user Caitlyn Wang
    When Sal used 8m/s in the beginning and in triangle formula....Where did he get it from?
    (2 votes)
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    • blobby green style avatar for user robshowsides
      Sal writes 8m/s for the first time at in the video. He started with the formula:
      vi + (delta t)(a) = vf
      5 + (4s)(2 m/s^2) = vf
      5 m/s + 8 m/s = vf
      13 m/s = vf
      So 8 m/s = (delta t)(a) = vf - vi = delta v
      This 8 m/s is the change in velocity (v increased by 2 m/s every second, for 4 seconds).
      That's why 8 m/s also became the height of the triangle, since the bottom of the triangle on the graph was at vi = 5 m/s, and the top of the triangle was at vf = 13 m/s. The graph (velocity vs time) is a straight line that rises with a constant slope of 2, because the velocity is increasing at a steady rate of 2 m/s (graph goes up 2) per second (graph goes right 1).
      (6 votes)
  • primosaur ultimate style avatar for user Bjorn.Frohlich
    I'm new here, sorry if I ask a stupid question.

    When Sal calculates the distance in the second half of the video. Wouldn't it be right to call it the displacement? My reasoning is that since he is calculating with velocity (a vector) and not just speed, the result wood be a vector again.

    (I am aware that in the given example distance and displacemet are the same).
    (2 votes)
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    • male robot hal style avatar for user Amiya
      Since, we are working with "constant acceleration" (magnitude and direction of acceleration will not change) motion will be in a straight line. So, we can use "displacement" and "distance" interchangeably.
      (2 votes)
  • leaf grey style avatar for user Bhargav Chirravuri
    Starting from Sal breaks the integral of the velocity-time curve into two parts. But wouldn't it be easier if he used the formula A(trapezoid) = 1/2(side 1 + side 2)*height, which in this case would be 1/2(v1 + v2)*deltaT? One step versus two steps.
    (2 votes)
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  • marcimus purple style avatar for user munozmarco24
    Do you also cancel units when you are multiplying? In this video, you cancel second (s) when you multiplied 4 seconds times 2 meters per second. I thought we only cancel units when we use division.
    (3 votes)
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  • aqualine tree style avatar for user Daniel Kaungzan
    What does Sal do in where he says that v(i) and 1/2 v(i) will simplify to 1/2 v(i)?
    (2 votes)
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    • blobby green style avatar for user robshowsides
      He is using algebra to simplify the expression in the parentheses. There is:
      ( v(i) + 1/2 v(f) - 1/2 v(i) )
      So he looks at the two v(i) terms and notes that x - 1/2 x = 1/2 x is always true.
      For example, 10 - (1/2)*10 = (1/2)*10, since 10 - 5 = 5. If you subtract half of something, you are always left with half of the original. So he has:
      v(i) - 1/2 v(i) = 1/2 v(i)
      and thus
      ( v(i) + 1/2 v(f) - 1/2 v(i) ) = 1/2 v(i) + 1/2 v(f)
      (3 votes)

Video transcript

The goal of this video is to explore some of the concepts, or formulas you might see in a traditional physics class. But even more importantly to see that they're really just common sense ideas. So let's just start with a simple example. Let's say that-- and for the sake of this video just so I don't have to keep saying this is the magnitude of the velocity, this is the direction of the velocity, et cetera, et cetera. Let's just assume that if I have a positive number that it means, for example, if I have a positive velocity, it means that I am going to the right. And let's say if I have a negative number, which we won't see in this video, let's assume that I'm going to the left. And that way I can just write a number down, are we're only operating in one dimension. You know that it is specifying what the magnitude and the direction is. If I said the velocity is five meters per second, that means five meters per second to the right. If I say it's negative five meters per second, that means it's five meters per second to the left. Now let's just say just for simplicity, let's just say we start with an initial velocity, of five meters per second. And once again, I am specifying both the magnitude and the direction. Because of this convention here, we know it is to the right. And let's say that we have a constant acceleration. We have a constant acceleration at play of two meters per second, per second, or two meters per second squared. And once again, since this is positive, it is to the right. And let's say that we do this for a duration. So my change in time is- let's say we do this for a duration of four-- I'll just use s. So I don't use seconds and then sec, and then s in different places. So s for this video is seconds. So I want to do is think about how far do we travel. Well there's two things. How fast are we going after four seconds, and how far have we traveled over the course of those four seconds. So let's draw ourselves a little bit of a diagram here. So this is my velocity axis and this over here is my time axis. I can draw a straighter line than that. So that is my time axis. This is velocity. That is my velocity right over there. And I'm starting off at five meters per second. So this is five meters per second right over here. So v i is equal to five meters per second. And then every second that goes by, I'm going two meters per second faster. Because it's two meters per second, per second. So every second that goes by-- so after one second, I'm going to go two meters per second faster. So I'll be at seven. Or another way to think about it, is the slope of this velocity line is my constant acceleration. Or I have a constant slope here. So it might look something look something like that. So what has happened after four seconds? So one, two, three, four. So this is my delta t. So my final velocity is going to be right over there. I'll write it down here just because it's getting in the way of this word velocity. And so this is v. This is my final velocity. And what would it be? Well I'm starting at five meters per second. So I'm going to do it both, using the variables, and using the concrete numbers. So starting at some initial velocity. The subscript i says i for initial. And then each second that goes by, I'm getting this much faster. So if I want to see how much faster have I gone, I multiply the number of seconds-- I'll just multiply the number of seconds that go by, times my acceleration. And once again, this right here, I just wrote the subscript c saying that it's a constant acceleration. And so that will tell me how fast I've gone. If I start at this point, and I multiply the duration times my slope, I will get this high. I will get to my final velocity. And just to make it clear with the numbers, these numbers really could be anything, I'm just picking these to make it concrete in your mind. You have five meters per second, plus four seconds, plus-- I want to do that in yellow-- plus four seconds times our acceleration of two meters per second squared. And what is this going to be equal to? You have the seconds canceling out with one of the seconds down here. So let me write it. So we have five meters per second, plus four times two is eight. This second's gone. We're just left with meters per second. Eight meters per second. Or this is the same thing as 13 meters per second, which is going to be our final velocity. And I want to take a pause here. And you can pause and think about yourself. This hopefully should be intuitive. We were starting at going five meters per second. Every second that goes by we're going to go two meters per second faster. So after one second we will be at seven meters per second. After two seconds we will be at nine meters per second. After three seconds we will be 11 meters per second. And then after four seconds we will be at 13 meters per second. So you multiply how much time passed times acceleration. This is how much faster, we're going to be going. If we're already going five meters per second. Five plus how much faster? 13 meters per second. So this right up here is 13 meters per second. So, I'll take a little pause here. Hopefully that's intuitive. And the whole point of that is to show you that this formula that you'll often see in many physics books-- It's not just something that randomly popped out of the air-- it just makes complete common sense. Now the next thing I want to talk about is what is the total distance that we would have traveled. Now we know from the last video that distance is just the area under this curve right over here. So it's just the area under this curve. You say wait, this is kind of a strange shape right here. How do I calculate its area? And we can just use a little simple geometry to break it down into two different areas, that are very easy to calculate their areas. Or two simple shapes. You can break it down into this blue part, this rectangle right over here. Easy to figure out the area of a rectangle. And we could break it down into this purple part. This triangle right here. Easy to figure out the area of a triangle. And that will be the total distance we travel. And even this will hopefully make some intuition. Because this blue area is how far would we have traveled if we were not accelerating. If we just went five meters per second, for four seconds. So if you go five meters per second times-- so this is one second, two second, three second, four second. So you're going from times zero to time four, your change in time is four seconds. So if you go five meters per second, for four seconds, you're going to go 20 meters. This right here is 20 meters. That is the area of this right here. Five times four. This purple or magenta area tells you how further than this, are you going, because you are accelerating, because you kept going faster and faster and faster. And it's pretty easy to calculate this area. The base here is still five, because that's five seconds have gone by. And what's the height here? The height here is my final velocity minus my initial velocity. Or it's the change in velocity due to the acceleration. And the change in velocity due to the acceleration. 13 minus five is eight. Or it's this 8 right over here. It is eight meters per second. So this height right over here is eight meters per second. The base right over here is four seconds. That's the time that past. So what's the area of this triangle? Area of a triangle is one half, times the base, which is four seconds, times the height, which is eight meters per second. Seconds cancel out. One half times four is two, times eight is equal to 16 meters. So the total distance we traveled is 20 plus 16, is 36 meters. That is the total, or I could say the total displacement, and once again, it's to the right, since it is positive. So this is our displacement. Now what I want to do is do this exact same calculation, but keep it in variable form and that'll give another formula that many people, many often memorize. But I want you to understand it it's a completely intuitive formula, and it just comes out of the logical flow of reasoning that we went through in this video. What is the area? Once again, if we just think about the variables. Well, the area of this rectangle right here is our initial velocity, times our change in time. So this is the blue rectangle right over here. And then plus-- what do we have to do-- we have the change in time, once again, we have the change in time, times this height. Which is our final velocity. Which is times our final velocity minus our initial velocity. These are all vectors. They're just positive, telling us we're going to the right. And we don't just multiply the base times the height. That would give us the area of this entire rectangle. We have to take it by half because a triangle is only half of that rectangle. So times one half. And so this is the area. This is the purple area. That's not purple. This is the purple area, right over here. This is the area of this. This is the area of that. And let's simplify this a little bit. Let's factor out a delta t. So if you factor out a delta t, you get delta t times, a bunch of stuff, v sub i. So your initial velocity. We factored this out. Plus this stuff. Plus this thing right over here. And we can distribute the one half. We factored the delta t's out. And let's multiply the one half times each of these things. So it's going to be plus, one half times v f, times our final velocity. That's not the right color, let me actually do it in the right color so you understand what I'm doing. So this is the one half. So plus one half times our final velocity. Minus one half times our initial velocity. I want to do that in blue. Sorry, I'm having trouble changing colors today. Minus one half times our initial velocity. And now what does this simplify to? We have something plus one half times something else, minus one half times that original something. So what is v i minus one half times v i. So anything minus half of it, you're just going to have a positive half left. So these two terms, this term and this term will simplify to one half v i. One half the initial velocity plus one half times the final velocity. And all of that is being multiplied by our change in time. Or the time that has gone by. And this tells us the distance. The distance that we travelled. Or another way to think about it-- let's factor out this one half. You get distance is equal to change in time, times-- factoring out the one half-- v i plus v f. No, that's not the right color. v i plus v f. So this is interesting. The distance we traveled is equal to one half of the initial velocity plus the final velocity. So this is really-- if you just took, this quantity right over here is just the arithmetic-- I always have trouble saying that word. It's the arithmetic mean of these two numbers. And so I'm going to define this as something new. I'm going to call this the average velocity. But we have to be very careful with this. This right here is the average velocity. But the only reason why I can just take the starting velocity and the ending velocity and, adding them together, and then divide by two. Essentially taking the average of these two things which would be someplace over here. And I take that as the average velocity, is because my acceleration is constant. Which is usually an assumption in most introductory physics classes. But it's not always the assumption. But if you do have a constant acceleration like this, you can assume that the average velocity is going to be the average of the initial velocity and the final velocity. If this was a curve or if the acceleration was changing, you could not do that. But what's useful about this-- is if you want to figure out the distance that was traveled you just need to know the initial velocity and the final velocity. Average the two, and then multiply that times the time that goes by. So in this situation, our final velocity is 13 meters per second. Our initial velocity was five meters per second. So you have 13 plus 5 is equal to 18. You divide that by 2. Your average velocity is 9 meters per second, if you take the average of 13 and 5. And then 9 meters per second times 4 seconds gives you 36 meters. So hopefully that doesn't confuse you. I just want to show you where some of these things that you'll see in your physics class, some of these formulas, why you shouldn't memorize them, and they can all be deduced.