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Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction. Created by Sal Khan.

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  • leaf red style avatar for user christian beck
    how do u find momentum
    (16 votes)
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  • leaf blue style avatar for user AstroPoet
    Ok so I just wanna make sure I understand this correctly: So the reason seconds is squared is because it is seconds PER second, right?
    (15 votes)
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    • blobby green style avatar for user hardy.alex11
      I think you get the point, but it may be easy to be misinterpret. If I were to try to figure out the number of units of time per another unit of time. It might be difficult to get from how it is said, exactly, but it's because you are taking velocity (which is already in meters per second) and dividing it further by another unit of seconds, meaning you effectively have ((m/s)/s).
      (6 votes)
  • leaf blue style avatar for user Sandra
    so speed without direction is a scalar quality and velocity is speed with direction making it a vector quality, right?
    (13 votes)
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  • aqualine ultimate style avatar for user Ηßτ∂nvεεr™
    As I know lots about cars, I know that accelerating in the beginning is easier than when you go higher because engine needs to maintain the old speed + add new speed (accelerate)......I am assuming the 20mph/second^2 to be the average of the three seconds of acceleration. For example, say that the car accelerated up to 23mph in the first second, 21 mph in the 2nd second, and 19mph in the 3rd second. It wouldn't have been exactly 20, 20, 20......So my question is, is there any way we can find the three possible exact values of each second of acceleration according to the info I have just given using physics?
    (12 votes)
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    • blobby green style avatar for user kevinpfromnm
      You're technically referring to average values for each second as well. From the initial set of information described, it is not possible to get back to more detailed information about each second or the more desirable instantaneous acceleration (the slope of the velocity graph at a particular point) from which it's possible to get any other information about the average acceleration overall or over any sub-period within said range (see integral calculus for details on how).

      Short of it, no. Averaging a set of data hides completely the details interior to it so you can't reverse it.
      (6 votes)
  • female robot grace style avatar for user Ankita
    At , how did he cancel out the hour from the denominator ?
    (3 votes)
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    • leaf green style avatar for user Antonio Martos
      To be honest, I find this explanation is confusing since it mixes incomplete explanations about acceleration and unit conversion.

      Your question is related to the second topic (unit conversion) and is not explained sufficiently here, I think. For some reason the lecturer is trying to put the final result in miles/second/second. Do not ask me why exactly, I would stick with miles/hour/second which are valid units, or go to meter/second/second (International units).

      Anyway, you have miles/hour/second and you want to change then hours to seconds.
      The way to do this is: You can always multiply by a fraction equal to 1 (which would change nothing). So your trivial multiplying factor would be 1 hour / 1 hour but since 3600 seconds equals to 1 hour, this can be written as well as 1 hour / 3600 which is a convenient conversion factor from hours to seconds. Then hour units cancel out when multiplying and you got your result in seconds.

      You may consider instead the reverse fraction: 3600 seconds / 1 hour, but you can discard this since it would not cancel out.

      This is general procedure for any units conversion and can be applied multiple times until converting to convenient units.
      (7 votes)
  • leaf green style avatar for user bezawitbelachew
    What about deceleration? what's the definition and how do you calculate it? is it any different from calculating acceleration? except for maybe some negative values?
    (4 votes)
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    • starky ultimate style avatar for user Brandon Gomes
      Deceleration is just acceleration, where the acceleration vector is pointing in the opposite direction in respect to the velocity vector. Usually in physics we only say acceleration but give a direction (because it is a vector).
      For example: if you are moving at 3m/s to the right and accelerating at -1m/s/s, then you are "decelerating" by 1 m/s every second. After 3 seconds you will have decelerated to 0 m/s.

      Do you understand? If not commment and we can continue the conversation.
      (10 votes)
  • blobby green style avatar for user 2691graceg
    I watched all the videos in this chapter but couldn’t find about the formula xf=xi+vit+1/2at^2 which was appeared at the position acceleration and velocity questions. Can I ask the explanation of that formula?
    (8 votes)
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    • piceratops ultimate style avatar for user idk
      If acceleration a is constant, it means that the velocity v is increasing linearly with respect to time (a = (vf- vi)/t or vf = vi + a. t), where vi is the initial velocity at time 0 and vf is the final velocity at time t . If velocity was fixed, then distance x will change linearly. But in this case velocity is changing linearly with time t, so the distance is changing quadratically (power of 2) with respect to time, then distance x = xi + b. t+ c. t ^2, where xi is initial distance at time = 0, b and c are constants. We need to solve and find b and c which are related to vf, vi and a .

      One way is to use differentiation. If we differentiate x with respect to t, we get velocity v = (change in x )/ (change in t ). x = xi + b. t + c. t ^2, after differentiation we get v = b + 2.c. t . comparing with the vf = vi + a . t, final velocity or velocity at time t is vf, b= vi , a = 2.c and c = a /2.

      Replace b and c in the equation, and distance x at time t is xt, we get
      xt = xi + vi . t + (a /2). t ^2

      This equation is for the distance x at time t, when initial distance is xi, initial velocity is vi, and acceleration is a .
      (1 vote)
  • blobby green style avatar for user micahS
    what is a porsche?
    (0 votes)
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  • duskpin ultimate style avatar for user Ozi Mbaekwe
    is acceleration always constant
    (3 votes)
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  • male robot hal style avatar for user Madd Sam
    I'm trying to understand what all of these kinematic formulas mean. So playing around with some graphs, I came up with a question. I am sure the answer is simple, but I don't yet know it.
    Imagine that I start with a velocity of 0 and accelerate for four seconds (constant), and have a final velocity of 3 m/s.
    Therefore, constant acceleration would = 3/4 m/s². The magnitude of the graph over this time period would = √3²+4².
    What does this magnitude represent, and how would it be used?
    (4 votes)
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Video transcript

In this video, I want to talk a little bit about acceleration. And this is probably an idea that you're somewhat familiar with, or at least you've heard the term used here or there. Acceleration is just the change in velocity over time. Probably one of the most typical examples of acceleration, if you're at all interested in cars, is that many times they will give you acceleration numbers, especially for sport cars, actually all cars if you look up in Consumer Reports, or wherever they give the stats on different cars. They'll tell you something like, I don't know, like a Porsche-- and I'm going to make up these numbers right over here. So let's say that we have a Porsche 911. They'll say that a Porsche 911, they'll literally measure it with a stopwatch, can go 0 to 60 miles per hour. And these aren't the exact numbers, although I think it's probably pretty close. 0 to 60 miles per hour in, let's say, 3 seconds. So, although officially what they're giving you right here are speeds, because they're only giving you magnitude and no direction, you can assume that it's in the same direction. I mean, We could say, 0 miles per hour to the east to 60 miles per hour to the east in 3 seconds. So what was the acceleration here? So I just told you the definition of acceleration. It's change in velocity over time. So the acceleration-- and once again acceleration is a vector quantity. You want to know not only how much is velocity changing over time, you also care about the direction. It also makes sense because velocity itself is a vector quantity. It needs magnitude and direction. So the acceleration here-- and we're just going to assume that we're going to the right, 0 miles per hour and 60 miles per hour to the right-- so it's going to be change in velocity. So let me just write it down with different notation just so you could familiarize yourself if you see it in the textbook this way. So change in velocity. This delta symbol right here just means "change in." Change in velocity over time. It's really, as I've mentioned in previous videos, it's really time is really a change in time. But we could just write time here. This 3 seconds is really change in time. It might have been, if you looked at your second hand, it might have been 5 seconds when it started, and then my 8 seconds when it stopped, so it took a total of 3 seconds. So time is really a change in seconds. But we'll just go with time right here, or just with a t. So what's our change in velocity? So our final velocity is 60 miles per hour. And our original velocity was 0 miles per hour. So it's 60 minus 0 miles per hour. And then, what is our time? What is our time over here? Well, our time is, or we could even say our change in time, our change in time is 3 seconds. So this gives us 20 miles per hour, per second. Let me write this down. So this becomes, this top part is 60. 60 divided by 3 is 20. So we get 20. But then the units are little bit strange. We have miles. Instead of writing MPH, I'm going to write miles per hour. That's the same thing as MPH. And then we also, in the denominator, right over here, have seconds. Which is a little bit strange. And as you'll see, the units for acceleration do seem a little bit strange. But if we think it through, it actually might make a little bit of sense. So miles per hour. And then we could either put seconds like this, or we could write per second. And let's just think about what this is saying, and then we could get it all into seconds, or we could all get into hours, whatever we like. This is saying that every second, this Porsche 911 can increase its velocity by 20 miles per hour. So its acceleration is 20 miles per hour, per second. And actually, we should include the direction, because we're talking about vector quantities. So this is to the east. So this is east, and then this is east right over here-- just so we make sure that we're dealing with vectors. You're giving it a direction, due east. So every second it can increase in velocity by 20 miles per hour. So hopefully, with the way I'm saying it, it makes a little bit of sense. 20 miles per hour, per second. That's exactly what this is talking about. Now we could also write it like this. This is the same thing as 20 miles per hour, because if you take something and you divide by seconds, that's the same thing as multiplying it by 1 over seconds. So that's miles per hour-seconds. And although this is correct, to me this makes a little less intuitive sense. This one literally says it. Every second, it's increasing in velocity by 20 miles per hour. 20 miles per hour increase in velocity per second. So that kind of makes sense to me. Here it's saying 20 miles per hour-seconds. So once again, it's not as intuitive. But we can make this so it's all in one unit of time, although you don't really have to. You can change this so that you get rid of maybe the hours in the denominator. And the best way to get rid of an hour in the denominator, is by multiplying it by something that has hours in the numerator. So hour and seconds. And here, the smaller unit is the seconds. So it's 3,600 seconds for every 1 hour. Or 1 hour is equal to 3,600 seconds. Or 1/3600 of an hour per second. All of those are legitimate ways to interpret this thing in magenta right over here. And then you multiply, do a little dimensional analysis. Hour cancels with hour. And then will be equal to 20/3600. 20/3600 miles per seconds times seconds. Or we could say, miles-- let me write it this way-- miles per seconds times seconds. Or we could say, miles per second-- I want to do that in another color-- miles per second squared. And we can simplify this a little bit. Divide the numerator and denominator by 10. You get 2/360. Or you could get, this is the same thing as, 1/180 miles per second squared. And I'll just abbreviate it like that. And once again, this 1/180 of a mile, how much is that? You might want to convert to feet. But the whole point in here is, I just wanted to show you that, well, one, how do you calculate acceleration? And give you a little bit of a sense of what it means. And once again, this right here, when you have seconds squared in the bottom of your units, it doesn't make a ton of sense. But we can rewrite it like this up here. This is 1/180 miles per second. And then we divide by seconds again, per second. Or maybe I can write like this, per second, where this whole thing is the numerator. So this makes a little bit more sense from an acceleration point of view. 1/180 miles per second, per second. Every second, this Porsche 911 is going to go 1/180 of a mile per second faster. And actually, it's probably more intuitive to stick to the miles per hour, because that's something that we have a little bit more sense on. And another way to visualize it. If you were to be driving that Porsche, and you were to look at the speedometer for that Porsche, and if the acceleration was constant-- it's actually not going to be completely constant-- and if you look at speedometer-- let me draw it. So this would be 10, 20, 30, 40, 50, 60. This is probably not what the speedometer for a Porsche looks like. This is probably more analogous to a small four cylinder car's speedometer. I suspect the Porsche's speedometer goes much beyond 60 miles per hour. But what you would see for something accelerating this fast is, right when you're starting, the speedometer would be right there. And that every second it would be 20 miles per hour faster. So after a second the speedometer would have moved this far. After another second the speedometer would have moved this far. And then after another second the speedometer would have moved that far. And the entire time you would have kind of been pasted to the back of your seat.