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Solving for time

Rate of change in position, or speed, is equal to distance traveled divided by time. To solve for time, divide the distance traveled by the rate. For example, if Cole drives his car 45 km per hour and travels a total of 225 km, then he traveled for 225/45 = 5 hours. Created by Sal Khan.

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  • leafers ultimate style avatar for user aroonim2000
    I don't understand how displacement can be negative, after all it is the shortest distance between the initial and final point ?
    (45 votes)
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    • duskpin seedling style avatar for user studioj.lardinois
      I'd like to add to this - I know this is an old question, but new readers might find this useful.

      The + and - designations are important in computer programming, especially with video game development.

      In video games, the motion of characters or objects and defined with vectors, just like this. However, in computer code, we have no way of cleanly saying "to the east," or "to the left."

      So instead, we will have something written like this: (assuming that "num" is a Vector object).

      num = (+1,0,0);

      This states that the vector called num is +1 on the X axis. If it were written as (0,-1,0), then this would denote that the object moves -1 on the Y axis, or what we might call "down".

      So the + and - don't mean that the actual value is negative, it just means that it is forward / backward relative to the * starting point * on the axis.

      So if the object is originally at (4, 2, 7), and the object moves, or is displaced, by a vector of (-1, 2, 0), then the final position is at (3, 4, 7).

      So + and - don't represent true positive and negative values, in this case, they just try to show whether or not a value on the X, Y, or Z access has increased or decreased from it's original spot / value.

      This is an important concept in video game development, and takes up a large portion of programming a typical video game.
      (66 votes)
  • duskpin sapling style avatar for user Ryan zebedee
    what is negative velocity
    (6 votes)
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  • marcimus pink style avatar for user Marium
    At , how did you manipulate the equations? Can you define it here please?
    (5 votes)
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  • leaf green style avatar for user ruthpoh99
    So can we solve this velocity question in the same way we solve a speed question? Is the only difference the addition of directions in this question?
    (10 votes)
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  • winston default style avatar for user FERDOUS SIDDIQUE
    Why do we use arrow on top of vector ?
    (6 votes)
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  • aqualine ultimate style avatar for user Moon_yl
    Why is the direction so important? Isn't the answer the same without a direction? Also, when you answer those physics questions, it's the working that gives you the mark is it?
    (2 votes)
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    • duskpin sapling style avatar for user E L
      You can view negative as distance travelled backwards and positive as distance travelled forwards. It just shows a clearer answer so that when you calculate for velocity which essentially requires displacement. You would not need the direction if you are calculating for scalar values.
      (2 votes)
  • winston baby style avatar for user Fatima Ghani
    How do these positives and negatives determine directions?
    (3 votes)
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    • starky ultimate style avatar for user Fantastic Flora
      It can determine directions such as + is going to the East, and - is going to the West.
      The positives and negatives just tell you if you are going with the original direction or the opposite direction. So if the problem asks how much you are going to the East, but you are actually going to the West, your answer would be negative.
      (5 votes)
  • primosaur seed style avatar for user Abhiram A.T.
    At 02.30 Sal multiplied the equation with t and then proceeded to do the rest. Is there a way to do it without multiplying it by t?
    (5 votes)
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    • male robot johnny style avatar for user Anthony Coelho
      Yes, instead of multiplying by time you can just plug the numbers into the equation:
      time = distance / rate (speed). In this case it would be:
      time = 720m / 3m per sec

      When you divide 720m by 3m/s the meters cancels out and you are then left with time which would be 240 seconds. This equation though is just a manipulation of
      rate = distance/time.
      (2 votes)
  • blobby green style avatar for user KIM NAEUN
    why we have to divise vector quantity and scalars quantity? what happened if we don't divise them?
    (3 votes)
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  • aqualine ultimate style avatar for user Kind Nesz
    Does time have vector 'version'? like speed vector version is velocity and distance vector version is displacement
    (3 votes)
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    • piceratops ultimate style avatar for user Hecretary Bird
      "Time" itself isn't a vector, just like position isn't a vector. You can give time a direction when talking about "elapsed time", or the difference between two points in time. This would be a vector quantity. We don't really need to consider time as a vector, though, because the direction won't change on you. Time always moves forward.
      (3 votes)

Video transcript

Let's work through another few scenarios involving displacement, velocity, and time, or distance, rate, and time. So over here we have, Ben is running at a constant velocity of three 3 meters per second to the east. And just as a review, this is a vector quantity. They're giving us the magnitude and the direction. If they just said 3 meters per second, then that would just be speed. So this is the magnitude, is 3 meters per second. And it is to the east. So they are giving us the direction. So this is a vector quantity. And that's why it's velocity instead of speed. How long will it take him to travel 720 meters. So let's just remind ourselves a few things. And I'll do it both with the vector version of it. And maybe they should say, how long will it take them to travel 720 meters to the east, to make sure, to make it clear, that it is a vector quantity. So that it's displacement, as opposed to just distance, but we'll do it both ways. So one way to think about it, if we think about just the scalar version of it, we said already that rate or speed is equal to the distance that you travel over some time. I might write t there. But it's really a change in time. So sometimes some people would write a little triangle, a delta there, which means change in time. But that's implicitly meant when you just write over time like that. So rate or speed is equal to distance divided by time. Now, if you know-- they're giving us in this problem, they're giving us the rate. If we think about the scalar part of it, they're telling us that that is 3 meters per second. And they're also telling us the time. Or, sorry, they're not telling us the time. They are telling us the distance, and they want us to figure out the time. So they tell us the distance is 720 meters. And so we just have to figure out the time. we So if we just do the scalar version of it, we're not dealing with velocity and displacement. We're dealing with the rate or speed and distance. So we have 3 meters per second is equal to 720 meters over some change in time. And so we can algebraically manipulate this. We can multiply both sides times time. Multiply time right over there. And then we could, if we all-- well, let's just take it one step at a time. So 3 meters per second times time is equal to 720 meters because the times on the right will cancel out right over there. And that makes sense, at least units-wise, because time is going to be in seconds, seconds cancel out the seconds in the denominator, so you'll just get meters. So that just makes sense there. So if you want to solve for time, you can divide both sides by 3 meters per second. And then the left side, they cancel out. On the right hand side, this is going to be equal to 720 divided by 3 times meters. That's meters in the numerator. And you had meters per second in the denominator. If you bring it out to the numerator, you take the inverse of this. So that's meters-- let me do the meters that was on top, let me do that in green. Let me color code it. So 720 meters. And now you're dividing by meters per second. That's the same thing as multiplying by the inverse, times seconds per meters. And so what you're going to get here, the meters are going to cancel out, and you'll get 720 divided by 3 seconds. So what is that? 720 divided by 3. 72 divided by 3 is 24. So this is going to be 240. This part right over here is going to be 240. And it's going to be 240 seconds. That's the only unit we're left with, and on the left hand side, we just had the time. So the time is 240 seconds. Sometimes you'll see it. And just to show you, in some physics classes, they'll show you all these formulas. But one thing I really want you to understand as we go through this journey together, is that all of those formulas are really just algebraic manipulations of each other. So you really shouldn't memorize any of them. You should always say, hey, that's just manipulating one of those other formulas that I got before. And even these formulas are, hopefully, reasonably common sense. And so you can start from very common sense things-- rate is distance divided by time-- and then just manipulate it to get other hopefully common sense things. So we could have done it here. So we could have multiplied both sides by time before we even put in the variables, and you would have gotten-- So if you multiplied both sides by time here, you would have got, on the right hand side, distance is equal to time times rate, or rate times time. And this is one of-- you'll often see this as kind of the formula for rate, or the formula for motion. So if we flip it around, you get distance is equal to rate times time. So these are all saying the same things. And then if you wanted to solve for time, you could divide both sides by rate, and you get distance divided by rate is equal to time. And that's exactly what we got. Distance divided by rate was equal to time. So if your distance is 720 meters, your rate is 3 meters per second, 720 meters divided by 3 meters per second will also give you a time of 240 seconds. If we wanted to do the exact same thing, but the vector version of it, just the notation will look a little bit different. And we want to keep track of the actual direction. So we could say we know that velocity-- and it is a vector quantity, so I put a little arrow on top. Velocity is the same thing as displacement. Let me pick a nice color for displacement-- blue. As displacement-- Now, remember, we use s for displacement. We don't want to use d because when you start doing calculus, especially vector calculus-- well, any type of calculus-- you use d for the derivative operator. If you don't know what that is, don't worry about it right now. But this right here, s is displacement. At least this is convention. You could kind of use anything, but this is what most people use. So if you don't want to get confused, or if you don't want to be confused when they use s, it's good to practice with it. So it's the displacement per time. So it's displacement divided by time. Sometimes, once again, you'll have displacement per change in time, which is really a little bit more correct. But I'll just go with the time right here because this is the convention that you see, at least in most beginning physics books. So once again, if we want to solve for time, you can multiply both sides by time. And you get-- this cancels out-- and I'll flip this around. Well, actually, I'll leave it like this. So you get displacement is equal to-- I can flip these around-- velocity times change in time, I should say. Or we could just say time just to keep things simple. And if you want to solve for time, you divide both sides by velocity. And then that gives you time is equal to displacement divided by velocity. And so we can apply that to this right over here. Our displacement is 720 meters to the east. So in this case, our time is equal to 720 meters to the east. 720 meters east is our displacement, and we want to divide that by the given velocity. Well, they give us the velocity of 3 meters per second per the east. And once again, 720 divided by 3 will give you 240. And then when you take meters in the numerator, and you divide by meters per second in the denominator, that's the same thing as multiplying by seconds per meter, those cancel out. And you are just left with seconds here. One note I want to give you. In the last few problems, I've been making vector quantities by saying to the east, or going north. And what you're going to see as we go into more complex problems-- and this is what you might see in typical physics classes, or typical books, is that you define a convention. That maybe you'll say, the positive direction, especially when we're just dealing with one dimension, whether you can either go forward or backwards, or left or right. We'll talk about other vector quantities when we can move in two or three dimensions. But they might take some convention, like positive means maybe you're moving to the east, and maybe negative means you're moving to the west. And so that way-- well, in the future, we'll see, the math will produce the results that we see a little bit better. So this would just be a positive 720 meters. This would be a positive 3 meters per second. And that implicitly tells us that that's the east. If it was negative, it would then be to the west. Something to think about. We're going to start exploring that a little bit more in future videos. And maybe we might say positive is up, negative is down, or who knows. There's different ways to define it when you're dealing in one dimension.