All right in this video I want to talk about position, velocity and acceleration. And just the relationship between these three and kind of the short answer is that the derivative of position tells you about velocity will give you an equation that describes velocity. If you take the derivative of an equation that describes your velocity, you'll get the acceleration, and I'll kind of digress at the end a little about that. But again, all that really I said, again. So a lot of times a little abbreviate.

Position with s of T. So you've got some function. S of T that describes your position. If you take the derivative of that function that's going to give you a new function that tells you about velocity and velocity is going to have a sign associated with it, either positive or negative. And again, you can think about the sign as saying, well, maybe either you're moving to the left it's negative or to the right if it's positive, or maybe you're moving away if it's positive and kind of, you know, back. Towards possibly leap, maybe you leave your home.

And if your velocity is positive that indicates you're moving away from home. If your velocity is negative, it indicates you're coming back to get speed is always positive. So all we really have to do is just take the absolute value velocity to get our speed. Well, if we calculate the instantaneous rate of change of velocity, well, when we take the derivative of velocity that tells us about acceleration, so the derivative of velocity, which would be the. Second derivative of our position function that's going to tell us about acceleration. So let's, possibly do kind of want a mechanical problem.

And then I'm just going to talk a little for a second. Why? This hopefully why this makes sense of displacement in meters of a particle moving in a straight line is given by the equation of motion. S, equals five T cube, plus three, T, plus eight, where T is measured in seconds. We want to find the velocity after two seconds. And then the acceleration after two.

Seconds as well. So all we have to do is just take some derivatives. So the derivative of our position function, and we can even stick a T in there.

If we want to well that's going to be the same thing as our velocity equation. Well, if we take the derivative of 5, T cubed, the x it would come down front. So we would have five times 3t squared. The constant just kind of comes along. The derivative of 3t is simply three and the derivative of positive. Eight is zero so we'll get. Fifteen T squared, plus three.

Well that means the velocity after well, two seconds all we have to do is just plug it in. So if we plug in two we'll get 15 times two squared, plus three 2 squared is four, so we'll get fifteen times, four, plus three, 15 times four, 60, plus three will give us 63. And it was in meters and seconds. Those were our units. So it says, your velocity is 63 meters per second. Okay. So nothing to do nothing tricky.

Nothing crazy other than taking a derivative and plugging in T equals two. You know, they could ask. Something like, when does your velocity equal 100 meters per second. So just to throw in something different.

When is your velocity equal to 100 meters per second? Well, in that case, you would just take your velocity equation. And now we would, we're trying to figure out winter velocity is 100. We would just plug 100 and on the left side. And then you would simply have to solve for T. So you could subtract 3 divided by 15 take square roots and do all that, so I'm not going to go through it, but that's. The. Difference, okay, you would just plug it in and then have to solve let's.

Do the original other problem I did say which was to find the acceleration? Well, again, we know our velocity here. Our velocity at time T is 15 T squared plus 3. If we take the derivative of velocity that is going to give us our acceleration function. And if we take the derivative of 15 T squared, well, the T squared will get 2t to the first the derivative of positive 3 is just going to give us Z.

And if we multiply 15 times 2t, hey, We get 30 T. So that says, our acceleration, after 2 seconds would be 30 times 2, or we would get 60 meters. And now in the denominator, we write seconds squared. Okay. So those are your units 60 meters per second squared should that be plural I, don't know, it's plural. So that's, the basic idea you're just taking derivative is all you're doing. And you know, so that's the short answer, if you start with position, take the derivative to get velocity, take the derivative of velocity to get. Acceleration, equivalently, if you start with position, take two derivatives to get acceleration, if you've seen anti derivatives, well, you just go backwards.

You know. So if you start with acceleration, the antiderivative is velocity. And then the antiderivative of that is going to tell you stuff about position. And even why does this make sense? You know. So again, a derivative, remember somehow, I derivative, a derivative quantifies and instantaneous quantifies and instantaneous rate of change.

Well, if you're. Sitting in your car, right, and you're driving along I'm going to try to draw a terrible little car, hey, that's, a great car. So if you're driving in your car right, you're here going down the road there, if you look at your speedometer, for example, you know, really that's telling you a speed because it's its it, technically doesn't, tell you a velocity when you look at your speedometer, right because it's, not telling you the direction that you're going its, just telling you, hey, you're going 50 miles, an. Hour or however, fast you're going it's a funny-looking speedometer.

So, however, fast you're going. It says, you're going 50 miles an hour. Well, again, what is that 50 miles an hour? Tell you it somehow quantifies the rate, the in sent Aeneas rate to the rate at that instant, when you look at the speedometer it's, telling you the rate at which your distance is changing. So again, if you have a function that tells you, you know, so maybe this is time T. And this is your distance, maybe from home.

Okay. So possibly. Your distance, possibly you're speeding up, because you want to get that, you know, get away from home really quick for some reason. Possibly you got a hot date. The idea is, if you calculate the slope of that tangent line it's going to calculate the instantaneous rate of change or again, it's going to tell you. The slope of this line is going to tell you the velocity. So one more time, you know, when you know when you're driving really fast, you don't say all to your friends, you don't say the rate at which.

Our position was changing was huge. You say, our velocity, or usually, you say, our speed, it was huge. Okay. We were going really fast our speed was really fast the rate at which our position was changing was really quick. So that's, the big use again, we talked about derivatives of it as being instantaneous rates of change. So depending on the situation, what does it mean to be an instantaneous rate of change? Well again, this is kind of the useful part of calculus as you know if you start with.

The physical situation, typically that derivative is going to tell you something, you know, some new physical information. And sometimes you have to think about what that new physical information represents as in this case. So, alright. So sorry for kind of waxing on here at the end again, the short answer just take derivatives. But again, to me, this is the useful part. I mean, this is a conceptual end of it.

And this is what makes it less mechanical. And this is what you know when you can, I think grasp. Some of this stuff, it becomes less symbol pushing, and it becomes more useful. So all right. Anyway, I hope this video helps. And if you have any comments or questions as always feel free to post them as comments.

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