 Nice hit! You know, life is much more interesting when velocity and position can change. I mean, think about the simple act of throwing a softball and hitting it. It's much more interesting if I'm able to take the ball from rest and get it moving to some velocity before the batter then has to move a bat from rest to some speed. The two connect and then the velocity of the ball can change. Acceleration makes life much more interesting. In this lecture, we're going to explore as many aspects of describing acceleration and looking at some kinds of constant acceleration as we possibly can. Just as we did with velocity, we're going to begin this part of the lecture video by learning to describe acceleration in a single dimension, that is, acceleration along only one coordinate axis. The position of an object is given by defining a coordinate system by applying a system of units to specify the location in that coordinate system. So for instance, in the previous lecture, we were thinking about a train moving along a track. It could go either forward or backward. That is almost a textbook definition of one-dimensional motion as long as we define the dimension as being forward along the track or backward along the track. And we can denote that the x-axis of our coordinate system. We can then specify where the train is considered to be at its origin and then all positions ahead of the origin are positive locations along the x-axis and all positions behind the origin are negative locations on the x-axis. We can use units like meters or kilometers, which is just a thousand-meter units, to then specify the location in that coordinate system, giving physical meaning to where one is on that coordinate axis. The velocity of an object, like the train, is merely the change in its position as time changes. And then from this, we were able to come up with a few ways of mathematically assessing velocity. If we want to characterize the velocity of an object over extended periods of time, we can define a quantity that's quite simple mathematically, and that is simply the average velocity. This is the change in the position from where the train is considered to have started to the final location where the train arrives after some time t. And you take the change in the position, the later position minus the initial position, and you divide by the later time minus the initial time. And that ratio is the average velocity. That time duration can be long or it can be short. But typically when we're talking about average velocities, we're not talking about infinitesimal little units of time. We're talking about long periods of time, macroscopic periods of time, seconds, minutes, hours, etc. Now the instantaneous velocity is a mathematical riff off of the average velocity. It's a deeper dive into the mathematics of what you mean when you say velocity. This is the average velocity, but in the limit that the time duration that you are considering approached an infinitesimal, that is delta t approaches zero in size. Now instantaneous here in a mathematical sense means average, but over so short a period of time that you don't really care about the duration. The duration size is really meaningless. It's as if you specified the velocity at exactly one and only one time because that time duration that you're considering is so short. And we saw in the previous example that if one graphs this, it is the slope of the line that relates position and time that tells you the velocity. The velocity is just the slope of the line. The instantaneous velocity is the slope of just a very narrow piece of the line. That line could be very complex. It could be a curve, for instance. Now it's natural at this point to go one step further and define the new concept. After all, if velocity is the change in spatial position during a change in time, we can then ask what is the change in velocity during that same change in time? What does that quantity represent? Velocity does not have to remain constant throughout the time period under consideration, and we saw that in the previous lecture. And the answer is acceleration. This is the concept we're going to explore in this lecture, and using acceleration we're going to take a slightly deeper dive into the same mathematical concepts that we explored in the lecture on position, time, and velocity. As you will see, its units, the units of acceleration, are meters per second squared. So they are meters per second per second. Velocity is meters per second, and if velocity changes in time, then we must have a change in meters per second per second or meters divided by seconds squared. This is the most compact way to represent this unit. Now, as was the case with velocity, there is average acceleration, which we will denote with a subscript AVG, A subscript AVG, and that is equal to the change in the velocity divided by the change in time. And then just as we did with velocity, we can define instantaneous acceleration as the average acceleration, but in the limit that that time slice that we're considering, delta t, is becoming smaller and smaller and smaller and smaller and smaller approaching zero. We will put this into practice in this lecture. So far we've really played around with the idea of an instantaneous quantity in a fairly abstract way, but we're going to give it concrete and definite form in this lecture, and you're going to see how to exercise this concept with a real situation. Let's take a closer look at acceleration. Now, here we're going to riff off the definition of instantaneous acceleration and dig a little bit deeper, look a little bit more closely at how to handle that mathematical definition with its funny limit of delta t goes to zero. And I promise you that this will be a deep dive, but it will be extremely rewarding because when you master the concepts that we will look at in this portion of the lecture, you will have taken the first steps into a much larger mathematical world, a mathematical world that as we will likely discuss in class, not only is important for physics but also for chemistry and biology, it's important for engineering, and it's important for a very hot topic these days, artificial intelligence, the heart of which is the mathematics that we are going to learn right now. So what do we really want when we want instantaneous quantities like instantaneous acceleration or instantaneous velocity? It doesn't matter which one, but we'll use acceleration as our example because that's the topic of this lecture. What we really want to know is the acceleration at a specific time, and let's call that time t. It could be 3 seconds, it could be 10 seconds, but we'll use a variable t to represent it. All we can really do, though, so far is calculate average acceleration, and we saw that that's defined as the ratio of the change in velocity divided by the change in time, where for average acceleration the change in time is typically considered to be a macroscopic or large-scale time, seconds, minutes, hours, days, something like that, not microseconds or picoseconds or femtoseconds or something like that, but when we're doing limits and sending delta t to zero, what we are in practice doing is considering smaller and smaller and smaller units of time and looking at the slope of the line that relates velocity and time to one another in those narrower and narrower chunks of time. So what we have to calculate is delta v, and we want to do this to get an instantaneous quantity for as tiny a step of time as we can imagine. So what this means is, you know, let's imagine that we can find the velocity using some kind of instrument. We can find the velocity at some time t. And then using that same instrument, we can find the velocity again at a slightly later time, t plus delta t, where delta t is some add-on to the original time that we were making our first measurement. And so we can just go ahead and compute this. We can compute the average velocity using the sampling of the velocity at time t and the sampling at the slightly later time, t plus a little delta t. So let's write that down. This is what that average velocity would look like mathematically. We have v as a function of time. That's what the parentheses represent. Velocity is a function of time. The later time is t plus delta t. So v of d plus delta t is really the velocity at that later time. And then we have v of t, which is the velocity that we measured at the earlier time. And we simply take the difference, and we divide by the time delta t over which this measurement was made, from the initial measurement to the later measurement. So this is just average acceleration. Now what we want to do is to compute that ratio in as short a delta t as possible. So we finally arrive at a firmer mathematical definition of the instantaneous acceleration. We are now going to take the limit as delta t goes to zero, and we're going to define the limit of this ratio as instantaneous acceleration, or simply acceleration. So whereas before, we had the limit of delta t goes to zero of delta v over delta t, now we see what the structure of that delta v is. That structure has now been revealed to us by making this thought experiment where we think about what it means to measure velocity at a slightly earlier time, t, and a slightly later time, t plus delta t. If we know this, either we know the form of this function or have the number, and we know this, for instance, again, having velocity as a function of time, a well-defined function of time, and if we know delta t, or if we can get rid of delta t somehow by expanding out the mathematics of this ratio and cancelling out as many delta t as we can, then we can finally take that limit, we can send delta t to zero, and we can get acceleration a. In fact, that is exactly what we are going to do in this lecture. We're going to look at a concrete example of doing exactly this. We're going to define the velocity at a later time, t plus delta t. We're going to define it at an earlier time, t. We are going to have a functional form for that. This will allow us to mathematically get rid of a bunch of delta t's, and then finally the ones that are left will send to zero, and will actually be left with a quantity that isn't zero at the end. It's remarkable. So let's do that. So here is the usefulness of this closer look that we just took. Why was this so useful? Because now, if we are given firm mathematical statements about how velocity changes with time, we can answer questions like the following. At time t equals three seconds, what is the instantaneous acceleration? Before, we just had an equation with nothing to plug into it. We had no way of answering this question, so we can actually take the first steps in addressing this question with a firm answer. So for example, we could be told that the velocity is a function of time, and that the functional form is what is shown here. That the velocity is equal to a constant, which I'll talk about in a moment, times a function of time, which in this case I have selected to be t squared. Now, what is b? Well, b is really anything we want it to be, a concrete number in here in a little bit, but I want to walk us through the mathematics of this first. b is just a constant, and it has units, such that when it multiplies t squared, you get something with units of velocity. What are the units of velocity? The units of velocity are meters per second. So if we have time squared on the right-hand side, that means we have seconds squared in the numerator, and that means that b must have seconds cubed in the denominator to make those seconds squared cancel out and give us just meters per second at the end of this. So check that. It's a simple algebraic exercise, good practice, good exercise at this point in the course to make sure that that makes sense. So feel free to pause the video here and verify that whatever b is, it must have units of meters divided by seconds cubed. Now that we have this information, we can start using this function in the definition of instantaneous acceleration to find the acceleration at any time t that is valid for this functional relationship here. So let's begin doing that. In the first step, we simply begin by writing down the equation for the instantaneous acceleration. a is the limit as delta t goes to zero of v at a time t plus delta t minus v at a time t, that whole quantity divided by the delta t. Well, that's all nice, but we actually have the function now for v. We know that v equals bt squared. So whatever appears inside the parentheses here and here, we will substitute into where you see t squared in this equation and we'll replace the v's of t's with an actual function, that is, bt squared. So let's do that. So if I do that, going one step into this process, I replace the v as a function of t with the actual function, b times the time quantity squared. Well, for the left term in the numerator, the time quantity is t plus delta t. So that whole thing gets squared. And then in the second term in the numerator, it's just v of t, so we have bt squared. So the t remains unchanged in this case. You don't have to add anything to it and then square it. It's just t. Now, if this step was a little uncomfortable, rewind the video, listen to the process one more time, try it yourself, and make sure that this step feels comfortable to you. Well, now that we have done this substitution, we see that we have an exercise in sort of high school algebra polynomial expansion, where we have a quantity, t plus delta t all squared. Well, we can actually write that out. We can expand that out. So let's do that. That's the next step. We expand out the square of t plus delta t. So that leaves us with a term distributing the b constant through the three terms that will wind up being in this expansion. We have a term that goes as bt squared. We have a term that goes as 2 bt delta t, and that's because we have two terms that are bt delta t and bt delta t, and you can just add them together. And then finally, we have a term that's b times the quantity delta t all squared. So that's the expansion of the polynomial, and then over here on the right, we still have that lingering second part of the numerator, the minus bt squared. Now, if you stare at this for a moment, you'll begin to realize that we can make one simplifying step forward, and that is to observe that we have a term negative bt squared, and we have a term positive bt squared, and they're added together. So we can cancel them out, and I will highlight them in red, and I will explicitly show you that this cancels with this, and that leaves us with just 2 bt delta t plus b delta t squared in the numerator. Now things get really interesting, because when we throw out those terms that cancel out, we immediately see that we have one more thing we can cancel out of this ratio, and that is delta t. Delta t appears as a multiplier of 2 bt, and it appears squared, next to the second term in the numerator, times b, and it appears once in the denominator. So you can make this step any way you like. I have chosen to simply notice that because the numerator contains the delta t here and the delta t squared here, we can cancel one delta t in each of these terms with the one delta t in the denominator, and I've highlighted those in red, so that completely removes this delta t, leaving just a number one behind, because delta t divided by delta t is just one, and over here we can only get rid of one of the delta t's that are squared. So the square goes away, but we're still left with delta t to the one power, which is just delta t. And finally this leaves us with a completely simplified looking thing, and this is the thing that we are now going to act taking the limit as delta t goes to zero. What do you observe about the argument of the limit function? The first term in the argument contains no delta t's, so nothing will happen to this term when delta t is sent to zero, but the second term contains a delta t multiplying the constant b. And so what will happen when delta t is sent to the value of zero in the limit of delta t goes to zero? If you said that the second term in this argument will itself also go to zero, you are absolutely correct. And that means that the final result after sending delta t to zero is simply that the acceleration a is equal to two times b times t. Two times a constant times your time variable. This is a concrete result. If somebody asked you at a time, t equals three seconds, what is the instantaneous acceleration? You are one constant away from being able to answer that absolutely concretely. We just need to know what the value of b is. Congratulations. Whether you have realized it or not, you have just done calculus. You have just done first semester derivative calculus. Here is the shortcut that you would learn in a calculus course. The activity that you have just completed on the previous slide is known as taking the first derivative of a function of time which we can generically denote f of t. The parentheses are translated into English as of the argument, so f being a function of t time. What was the function of time that we took the first derivative of? Well, f of t was v of t and v of t was bt squared. So our function is the velocity and the variable of which it is a function is time. So when we took the derivative, that is when we took the limit of the average acceleration for the case that delta t goes to zero, in the language of calculus, we were taking the derivative of the velocity with respect to time. That's what this quantity really represents. Acceleration is the first derivative with respect to time of v, and that is denoted by this more compact notation, dv dt. This thing, d over dt, is really itself known as an operator in mathematics. It is a quantity that acts on a function and returns a function. That is, for our purposes, what an operator is. The first derivative is an operator. You feed it a function, it returns a function. That's its duty. What is the mathematical form of the first derivative with respect to time? It's this thing we've been playing with. It's the limit as delta t goes to zero of the change in velocity over the change in time. Now for our specific function, we can simply write out what the first derivative was. The acceleration was the first derivative with respect to time, d over dt, of the function velocity, which is itself a function of time. Plugging in that function, that was bt squared, and we got the answer that when we did all this out using all the steps on the previous slide, we got the answer to bt. Two times b times t. You've just taken your first time derivative whether you realized it or not. There are many more time derivatives ahead of you, so if this was uncomfortable to you, go back, practice it one more time, pick a different function, try using a different function, try using bt cubed, or bt to the fourth, and see what you can do with that. Now, it's important to note that if you stare at this page closely, you'll see the shortcut. There is a nice shortcut that you learn when you do calculus for the first time, when you're learning the derivative and you are learning the tricks of calculus to make the derivative go more quickly when you have to do it, say on a homework assignment or on an exam or in the real world in an engineering problem or in a chemical gradient problem, something like that. There is a shortcut that gets you to the functional answer much faster. To take the first derivative of a simple function like this one, where we have a constant times time to some power, all you have to do is simply multiply the function by the power of t and then reduce the power of t by one unit. So take a look at this. Here, the function was bt squared and I can tell you that if you use what I just said here a moment ago, multiply that function, bt squared, by the power of t, and then reduce the power of t by one unit, you will get to the answer that we got to after all those steps on the previous slide. This is much faster. So here's the shortcut. Take the power of t to multiply this whole function by two, so you would wind up getting two bt squared. Now reduce the power of t by one unit, which then gives us two bt to the one or just two bt. And that indeed was the answer that we got. So now you see what calculus really is. It is considering quantities where we send aspects of that quantity to, for instance, zero size and evaluate the pieces that are left over, which may not in fact be zero at the end. This is just one example, but there are many more that are going to come up in this class. So this is definitely a good couple of slides to practice or review many times during the course. The power of calculus cannot be understated. In fact, as a branch of mathematics, it was invented in the form we know it today, nearly simultaneously by two brilliant individuals. They did this development within about ten years of one another. And those two individuals were Sir Isaac Newton, who we will learn more about as we go through this course, and Gottfried Leibniz. Using this mathematics, we contain, as they needed to, tame very complex problems. And in fact, from the perspective of Isaac Newton, the reason that he needed to invent calculus in the first place was because it was impossible to describe in a simple way the observations of the natural world that he and his predecessors had made up to that point without this new kind of mathematics that considers quantities in the limit that they go to very small values, and then what happens to the functions of those quantities as a result of that. Isaac Newton had a wealth of information about planetary motion, about the motion of the moon and the sky, about objects falling to the ground on earth, about objects rolling, sliding down hills, etc. But in order to make sense of the natural laws that underlie all of those phenomena, he needed a mathematics that would allow him to describe changes as a result of some variable. And that is what we now call the calculus or simply calculus. And you see how this is extremely powerful. Given a function that relates velocity and time, you are now able to compute the acceleration at any time during the motion. And so for instance, using the previous example where we had velocity as a function of time being given by bt squared, we can now define the constant. We can now define b to be 2 meters per second cubed. And given the original question, what's the acceleration at t equals 3 seconds, we can answer the question. We know that acceleration equals 2 times b times t. And if we plug in b equals 2 with units of meters per second cubed, and t equals 3 with units of seconds, we can then do the remaining mathematics at this point, multiply 2 times 2 times 3, which is 12. And the final answer is 12 meters per second squared. So we see, checking our answer, that we have a number with the correct units for acceleration, meters per second squared. I mentioned earlier that you would see how the units of acceleration are meters per second squared. This problem demonstrates this quite nicely. But we could have gotten to this conclusion simply by looking at the definition of average acceleration, d meters per second divided by delta t, which is seconds. So that must be meters per second per second, or meters per second squared. And indeed, checking our answer here, we have a number with units of meters per second squared. So we can at least be assured we got the units worked out correctly. And that when we did the derivative, that is when we took the limit of the average acceleration in the limit that delta t goes to zero, indeed we arrived at a quantity that has the correct units for acceleration. Now, let's consider a special case of acceleration, and that is when acceleration is constant. In that case, it would mean that the ratio of delta v1 and delta t1 in one time duration is identical to the ratio of delta v2 and delta t2 in any other time duration. Okay? Now, I've written this out here in a sort of mathematical way. The acceleration in some time duration one would be given by this ratio, delta v1 over delta t1. And what we're saying is that under the special case that in any other delta t, so for instance some delta t2 elsewhere in the motion, the delta v divided by delta t in that unit of time gives you the same number, let's call it a2, then we can say that the acceleration is some constant. So, all of these are equal to each other under that special case. And we can represent this graphically using the same techniques that we did for position versus time, now for velocity versus time. So over here on the right I have a graph of velocity on the y-axis and time on the x-axis or the vertical axis and the horizontal axis depending on how you want to say it. And the arrow, the black arrow represents the relationship between velocity and time and if we slice the velocity at two different time segments, so for instance here is delta t1, the first of the two segments in the wide green rectangle and here is delta t2, the second of the two time segments in the second rectangle. We can look at specifically the slope of the line inside of those segments here and here and ask questions like is the slope changing? After all you would be getting a, for instance a different delta v over delta t in those two time slices. So what you can see in this graph here however is that the slope of the velocity versus time relationship is the same in both time slices. I could imagine taking this little line right here and just transporting it down here and you'd see that it lays right on top of the line inside of this green rectangle. So it must be the case that the slope is the same and since the slope of this line is the acceleration it must be the case that the acceleration is the same. So the acceleration can be said to be constant and in fact this kind of graph that you see here is exactly what happens when acceleration is constant. It's worth noting that this same argument can be applied back to constant velocity. That is if the slope of the change in position with respect to the change in time is the same in any two durations of time it must be the case that the velocity is constant if the slope is the same. Just as here because the slope of the velocity versus time is the same in the two units. It must be the case that the acceleration is the same. Now the beneficial thing about considering constant acceleration is that when you have that situation there are some extremely handy equations. We call equations of motion and they relate spatial location, time, velocity and acceleration assuming that the acceleration is constant with time. If acceleration is not constant with time these equations do not necessarily apply in the form that you will see written here. So we're going to look at these equations. We're not going to go into any depth about them right now. We're going to use class time. You should use the reading to expand your depth and understanding of these equations and class time and homework will be your first practice with utilizing them to answer questions about the physical world. But here I will merely introduce them and we will also look at some new mathematical terminology that appears in these equations. It's fairly straightforward terminology to grasp once you see the pattern. So the first equation is simply that the velocity at some time t is equal to the initial velocity plus the acceleration times the time that has passed. It's always assumed that the motion began at time zero which is why you don't see a t with a subscript zero appearing in these equations although you very well could put that in here. This equation is very useful if for instance you know the initial velocity you know the constant acceleration and you want to find the velocity at any other time t after time zero. It's also useful if you know the final velocity but don't know the initial velocity but have the acceleration and the time and you can solve for your unknowns. If as long as you only have one unknown then one equation will do just fine. But if you have more than one unknown you need more than one unique equation in order to solve for the unknowns. Two unknowns you need two unique equations. Three unknowns you need three unique equations and so forth. This should remind you of high school algebra where you should have learned to solve systems of equations with one or more unknowns that needed to be determined. The new terminology introduced in this equation is the v with a subscript zero which merely refers to the object's velocity at time t equals zero. And you can see that for instance if we set t explicitly to zero in this equation v is equal to v with a subscript zero. The next equation relates velocity acceleration and position. The first one related velocity acceleration and time. So now we see that we have the final velocity squared is equal to the initial velocity at time t equals zero squared plus a term that is two times the constant acceleration times the displacement in space during that time. X minus X naught. An X sub zero or X naught is simply the object's position at t equals zero. You may hear this kind of mathematical notation stated in various ways. The most common are X sub zero or X subscript zero or X naught which is n a u g h t for zero. X naught is typically how I will refer this or v naught. The next equation we have is the displacement X minus X naught is equal to the initial velocity times time plus a term that is one half the constant acceleration times the time squared. The next equation is that the displacement is equal to one half the initial velocity plus the final velocity. That whole quantity you're going to divide by two and that is multiplied by a half and that whole thing is multiplied by the time. And then finally we have that the displacement is equal to the final velocity at time t multiplied by that time minus one half the acceleration times time squared. You'll note that this equation is very similar to the third one but with the key difference that here we have the initial velocity and so if the acceleration is positive we would expect this to displace the object further forward along the X axis thus the plus sign. But here if we've displaced in the forward direction along the X axis and we only know the final velocity we need to subtract this term to get back to the sort of initial conditions. Anyway you'll have lots of opportunities to practice this beginning in class more so on the homework to give you some level of mastery and of course we will use and reuse these equations throughout the entire course to relate space, time, velocity and acceleration assuming that the acceleration is constant with time. And that's what I want to emphasize here at the end of this part of the lecture. It's important to remember that at all times the acceleration A is constant and the velocity V and the displacement X are the velocity and position as measured at time t. Now let's consider a force that yields a constant acceleration. This is a great way to employ these equations of motion to begin tackling problems in the natural world right away. And that case is gravity near the Earth's surface. Gravity as a force close to the Earth's surface is an excellent example of a constant accelerating force. If you take an object and you drop it in the gravitational field very close to the surface of the Earth, regardless of the mass of the object what you will find is that the acceleration experienced by those objects is always the same. Now we're going to study gravity in some detail later in the course and you'll see why it is that at least in this part of the physics curriculum, the mass of the object doesn't matter that the acceleration felt by any two objects is the same close to the surface of the Earth. But for now it's simply sufficient to know that if you take an object and you hold it out and you drop it near the surface of the Earth it will accelerate toward the center of the Earth that is it will head toward the floor or the ground and the acceleration that you are witnessing as it falls is nearly uniform and constant and independent of the mass of the object you drop. Now the reason I have a star here is I'd like to put a little caveat here. Of course when you drop an object close to the surface you are not dropping it in empty space with no other matter in it you are surrounded by the atmosphere, by air, nitrogen oxygen, etc. Air exerts a little bit of a resistance on the motion of a moving object and as we will see later when we explore forces that dragging force that air exerts always opposes the motion of an object. If you try to move an object to the left the force of the air resistance will push to the right. If you try to move to the right it will push to the left. So it is certainly true that when you drop objects near the surface of the Earth the shape of the object combined with its mass may change the amount of air drag that the object experiences as it accelerates toward the ground toward the center of the Earth. But if you can concoct a situation where that drag force is negligible then you can witness that the reality that the acceleration due to gravity is independent of the mass. So this star here is the caveat that yes there is air resistance but we will try to consider situations where it can be neglected for the time being so that we can make progress in understanding the natural world. Of course we can incorporate it later once we start thinking about forces. Now this acceleration this constant acceleration near the surface of the Earth is a useful number to memorize. It is 9.8 meters per second squared and this number is so useful that it is actually given its own symbol little g and it is referred to as the acceleration due to gravity near the surface of the Earth or simply g. And g is 9.8 meters per second squared. We can reveal the character of the constant accelerating force of gravity near the surface of the Earth by dropping something like a tennis ball. The video physics software allows us to track the motion of the tennis ball as it falls. Now in the previous video the motion was slowed down artificially using Apple iOS's slow motion video capability but for the video physics software analysis we want to return the video speed to normal and so the analysis that follows in this lecture is done using the video at normal speed. The video physics software allows you to track the motion of the tennis ball and what the software does as you can see in this still image is that it places a dot at each location in the video as the video advances where it thinks the tennis ball can be found. Now I didn't do any precise estimates when I dropped the ball in this video but based on where the ball starts up near my forehead and where it hits on the floor and knowing my height I would estimate that the ball fell a distance of about 1.73 to 1.77 meters or so. Something in that range. Using that information and using the graphs that you can obtain from the video physics software it's possible to get an estimate of the acceleration due to gravity that the tennis ball experiences. This is one of the graphs that the video physics software can return to you using the data that it obtains from frame by frame analysis of the video tracking an object frame by frame and knowing the time of each frame in the video. In the top graph we have a reporting of the Y coordinate of the ball as it falls during the time interval of the video. The video begins at time 0 but the ball is not dropped until about 4 tenths to 5 tenths of a second into the video. At that point once the ball is released by my hand you'll notice that it's Y coordinate changes and it does so more rapidly with each interval of time. The slope of the Y coordinate graph versus time is not constant. It changes in time. Now using delta Y divided by delta T in each time interval the software then converts this distance versus time information into velocity versus time information and so here we can see it was roughly this point at about 0.45 seconds or so into the video. It was at this point that the ball was released and began its descent to the floor and what we observe is that while the Y position, the vertical position versus time has a curve that has a changing slope with time the slope of the velocity versus time graph does not change of course until the ball hits the floor and then it bounces and then it changes quite rapidly and we see it rebound as the velocity changes in the graph. But let's only focus on this portion of the graph that I'm highlighting here with the arrow from the time when the ball is released from my hand to about the time where it hits the floor. It goes slightly off the camera frame when it hits the floor and the most reliable data point for the last part of the motion is probably somewhere back here around 0.93 seconds into the video. So our first data point with a velocity of about 0 is at around 0.45 seconds and our last data point that I would consider reliable while the ball is still essentially in view is somewhere around 0.95 seconds and that at this point the ball is moving in the negative direction it's moving down toward the ground with a velocity of about negative 5.5 meters per second and if you take delta V and divide by delta T using the numbers I just gave you you'll find out that the acceleration due to gravity over this period was on average about negative 10 meters per second squared and given the rough estimate of the ball's height started from and how far it dropped that's pretty good considering the true answer is around 9.8 I would assume I'm accurate with my height measurement to know more than about 5 to 10% and we've certainly gotten to within the ballpark of 9.8 meters per second squared for the acceleration due to gravity. Now is that acceleration constant on the tennis ball? Well it sure looks like it but the graph here while it's not a perfectly straight line and that can be attributed to uncertainty in the measurement from the frame by frame video analysis is pretty close to a straight line on average all the way down until the ball goes out of the frame of the camera and hits the floor so we see here on the tennis ball that there definitely is a constant acceleration of 9.8 meters per second squared I promised you that gravity is a force of acceleration near the surface of the earth that is constant and independent of the mass of the object that is dropped let us demonstrate that I have here a hammer and I have the tennis ball the hammer weighs much more than the tennis ball it feels much heavier in the hand than a tennis ball does but if I drop them starting from exactly the same height the hammer held upside down is it just about the same height as the base of the tennis ball and let them go we can see what happens indeed the head of the hammer and the bottom of the tennis ball strike the floor at just about the same time despite the fact that the hammer is much much heavier than the tennis ball indeed objects near the surface of the earth dropped in the gravitational field of the earth and neglecting air resistance experience exactly the same acceleration a constant acceleration of 9.8 meters per second squared the key ideas we have learned in this section of the course are as follows we have learned to define acceleration and we have seen that the average acceleration and the instantaneous acceleration have been defined using identical techniques to those same average and instantaneous concepts in velocity but in this part of the course we have taken a deeper dive into instantaneous quantities our practice, our example and we have seen how what emerges from considering instantaneous acceleration is a glimpse at the power of the calculus or just calculus which is extraordinarily useful in handling complex mathematical situations we have learned how to define the first derivative with respect to time of a function of time the function of time we have considered is velocity we have learned about constant acceleration to relate position, velocity, time and acceleration under that condition at least we have seen the equations that do it you will get more practice and insight from the reading, from class and from homework and we have learned that the earth's gravitational force close to the earth's surface is approximately a situation of constant acceleration that is independent of the mass of the object experiencing the gravitational attraction and this concludes this lecture on acceleration and concepts related to acceleration in motion