In this video we discuss displacement, velocity, and acceleration which are important specialized functions that use the quantity time as an input and have familiar physical interpretations. The relationships between displacement and velocity, and between velocity and acceleration serve as prototypes for forming derivatives, the main theme of this module, and towards which we'll develop formal definitions in later videos. Displacement functions describe the position or distance an object has moved at any particular time. The last example in the previous video involved a displacement function, where we looked at distance traveled between Sydney and Melbourne as a function of time over an interval of about 11.5 hours. In that example we looked at the average rate of change which gave us the average velocity for the entire journey being the slope of the line that joins the end points of the curve. We also introduced without too much formalism the idea of an instantaneous speed, which is what you see on the car's speedometer at any moment, and which is represented by the slope of a miniature tangent line attached to a point on the displacement curve. There are several places where the average speed for the whole trip must have appeared instantaneously on the car speedometer. An example so far we follow conventions in using x to denote the input and y to denote the output. It's common however when working with displacement functions to use t for time as the input variable, and x to denote displacement, the output variable. This is one of the few instances in which we would use the symbol x as a dependent rather than as an independent variable. Consider a cannon positioned to fire directly upwards, and use an x-axis now to make measurements in the vertical direction. The cannonball is fired and we watch its trajectory over six seconds. Its movement is subject to gravity so the ball goes up and then comes down. We'll ignore things like air resistance. Here are all the snapshots spread out evenly on the page. We regard the aperture of the cannon at height zero meters, and draw horizontal axis that denotes time passing. We also reposition the vertical axis, so that the moment the cannon fires is regarded as time zero seconds. Here's a table of values indicating the height x meters of the cannonball at time t seconds. Joining the points with a smooth curve in fact produces an inverted parabola, and there is a very good reason for this which we'll come to. Now the cannonball is slowing down under gravity but it's still rising for the first three seconds. We can work out the average velocity over these seconds, the slope of the line joining the points of the curve for t equals zero and three, and the answer turns out to be 17 meters per second. During the last three seconds, the cannonball just reaches its highest point and then is hurtling back down towards the earth. The average velocity over these seconds turns out to be -13 meters per second. Now that the answer is negative because the cannonball is moving downwards and we're using upwards as positive. I was careful to use the word velocity rather than speed because velocity distinguishes between positive and negative directions. Now in fact, the table of values comes from an underlying formula which happens to be a quadratic function, x equals minus five t squared plus 32 t plus three. This turns out to be quite a good model of the physical reality. Now imagine there's a tiny speedometer on or inside the cannonball, so that a tiny passenger can read the instantaneous velocity. The average velocity over the first three seconds was 17 meters per second. We can locate a miniature tangent line that just touches the curve and at that moment corresponding to that point the tiny speedometer would show 17 meters per second. Similarly for the last three seconds there'll be some moment when the cannonball is instantaneously traveling back to Earth at 13 meters per second. Now this works in general. Instantaneous velocities are represented by slopes of tangent lines anywhere on the curve, and of course the slopes can vary. For example, if we focus on this special point where the tangent line is horizontal, then the slope is zero. So, the instantaneous velocity is zero, which is the moment where the cannonball stops moving. So it's at that the point of returning to Earth. This corresponds to the apex of the parabola representing the displacement function. Because we have in fact a rule for the function which turns out to be a quadratic, we can use the trick of completing the square to find out exactly where this apex occurs. You can carefully check the working. Here's the rule for x, now expressed in a very useful form. So, we can read off the maximum which is 271 divided by five or 54.2. Which occurs when the square is zero, that is t is equal to 32 divided by 10 or 3.2. Thus, after exactly 3.2 seconds have elapsed the cannonball reaches a maximum height of 54.2 meters. Now we can go back to the original data and the graph that joins up the points, and investigate the velocity of the cannonball as a function of time t denoted by v. We can expand the table by adding a third row for v and try to figure out its values. That is, the instantaneous velocities of the cannonball at each of these points in time. Remember the instantaneous velocity is the slope of the tangent to the curve at that point. We can in principle use the diagram to find these slopes. For time t equals zero, the slope turns out to be 32, so v is 32 meters per second. For t equals one we find that the slope is 22, so v is 22 meters per second. And so on for t equals two, and t equals three. For t equals four, five, and six the slopes are negative and increasing in magnitude as the cannonball moves faster and faster as it falls back to Earth. Here's the displacement curve which is a parabola. Let's plot the points for velocity from the table and join the points to form a smooth curve which happens to be a straight line. In fact v is equal to negative 10t plus 32. This follows from the data in the diagram but later you'll learn a technique that enables you to deduce this algebraically from first principles. We can ask how the velocity changes, and in fact it decreases uniformly with time. Now acceleration is the technical term used to capture the rate of change of velocity. In this example, the acceleration turns out to be negative and constant. Negative acceleration is also called a deceleration. Write a for acceleration as a function of time t. Now we can add the graph for acceleration and it's described by the constant function a equals minus 10. In this example we have a quadratic for displacement, a linear function for velocity, and a constant function for acceleration. When we've developed some tools of calculus you'll be able to derive the rule for v and a from the rule for x very quickly. Also armed with integration techniques from the last module of this course given this rule for a you'll be able to work backwards to get the rule for v and x. This is where the physics comes in. In fact, the deceleration of an object close to the earth's service ignoring things like air resistance may be regarded as constant and it's about 9.8 meters per second per second. This is close to 10 meters per second per second used in this example. So, what we described turns out to be a reasonable approximation of the true behavior of this cannonball. What we did in this example generalizes. Finding slopes of tangent lines takes displacement to velocity, and this also in turn takes velocity to acceleration. Our cannonball example was rather special in that the velocity and acceleration we're a lot simpler than the displacement, but weird things can happen and you might ask generally what lies beyond acceleration? Consider our cannonball again but suppose it's attached by very strong spring at the origin and bounces up and down vertically between zero, one, zero, minus one and zero just like the values of the sine function. What you're witnessing is something known as simple harmonic motion. If you plot the displacement curve, you'll discover it's in the shape of a sine curve. Its velocity curve you'll discover looks like a cosine curve. It's acceleration curve looks like the sine curve tipped upside down. So, what comes next? What describes the instantaneous rate of change in acceleration? Engineers use the word jerk. Acceleration is related to forces acting on bodies. So, if there are sudden changes in acceleration, then the associated forces change rapidly, as in the case with extreme rides in amusement parks and there's much toing and froing, or jerking. In this example, the jerk curve turns out to be the velocity curve upside down. So, what comes next? What describes the instantaneous rate of change in jerk? The technical term is jounce. Amazingly in our example this becomes the original displacement function x equals x of t. Jounce is also called snap, so snap reproduces the original displacement curve. What's the instantaneous rate of change in snap? Physicists call it crackle. In the case of simple harmonic motion, crackle in fact coincides with velocity. So, what's the instantaneous rate of change in crackle? Physicists call it pop and this coincides with acceleration. We're now cycling through our original displacement, velocity, and acceleration and can continue reproducing these curves at ad infinitum just by taking instantaneous rates of change, a truly remarkable phenomenon. So, today we've looked in some detail at relationships between displacement, velocity, and acceleration and even mentioned briefly what happens beyond these. A direct link between displacement and velocity, between velocity and acceleration is the concept of a derivative, which were steadily working towards and which formalizes the technique and finding the slope of the tangent line at any given point on the curve. Please read the notes and when you're ready please attempt the exercises. Thank you very much for watching and I look forward to seeing you again soon.
