 Let's look at an example of a rectilinear motion problem. In this case we have a particle that moves along in line according to the equation s of t is equal to 2t cubed minus 9t squared plus 12t minus 4. Let's take a look at the graph of that so we can get a sense of the path that this particle is following. It is, of course, a cubic equation. I already have it set up here under y equals. So let's take a look at the graph. I did adjust the graph. If you care to try this on your own, you can see what I used. So the way in which I have this scaled is zero is over here at the far left t equals zero and it's 1, 2, 3, 4 over at the far right. So it is definitely a cubic equation. So let's go back and take a look at the questions we are asked to solve. So the first thing we want to determine is the time values for which the distance of this particle is increasing. Now remember the rules that we have. The distance s is going to be increasing when velocity is greater than zero. So the velocity we obtain by taking the derivative of the position function, establishing this relationship right here, is very important. So be sure to do that. So that gives us 6t squared minus 18t plus 12 and we want to see where that is greater than zero. So one easy way to do that is to solve it graphically. So we're going to go back to the graphing calculator and we are going to graph that equation. So please do try this along with me. Under y2 you'll put your derivative, your velocity expression, and we'll go ahead and just use the same graph. So if you happen to have the color version of the TI-84, this is the red graph. So we want to see where this graph, the red one, which represents the velocity, is greater than zero. So where does this graph lie above the x-axis? Well, it's crossing here at one and here at two and for values less than one, so it looks like in between zero and one, we have that the position is greater, or the velocity is greater than zero, and then from two to infinity. So those would be our answers for this question. So we have that the position s for this particle is increasing when time is in between zero and one and then from two to infinity. Let's consider another question. Now we want to find the t-values for which the velocity is increasing. Once again, remember our rules. Our velocity is increasing when the acceleration is greater than zero. So our acceleration is of course the derivative of velocity again. Establishing that relationship is very important. That's going to be 12t minus 18, which is very easy to solve that inequality. We get that t is greater than one and a half, three halves. So any time your time value is greater than one and a half, that is when the velocity will be increasing. Let's take a look at another question. Now we want to find the time values for which the speed is increasing. We have very specific rules for this. Remember speed is going to be increasing when the velocity and acceleration are both greater than zero or when velocity and acceleration are both less than zero. So we essentially have two problems to solve and we can solve this graphically as well. We are going to go back to our graph and calculator. We already have the velocity graphed. We are going to go ahead and graph that acceleration, 12t minus 18 and then answer these questions from that. So under y3, we will have 12x minus 18 representative of the acceleration. If we go ahead and graph that, remember the red graph was the velocity and the black graph here is the acceleration. So let's first consider where both velocity and acceleration are greater than zero. So we're talking about both the red graph and the black graph, both lying above the x-axis at the same time. Now it doesn't look like that happens till over here at 2 because over here at 2 you can see my pointer here. That's where the red graph crosses the x-axis after which it is greater than zero. Lying above the x-axis and at that same place you can see the black graph representing the acceleration is in the first quadrant there greater than zero as well. So the answer for this part will be from 2 to infinity. Now let's consider where both the velocity and acceleration are less than zero at the same time. So back to the graph. So now we need both the red graph and the black graph to be under the x-axis. So the red graph, the velocity, starts being less than zero right here where t equals 1 and let's see, we have to find where this black graph right here crosses the x-axis. I'm guessing it's one and a half, but let's go ahead and try it. We need to make sure we're on the correct graph and I need to find this x-intercept right about there. So let's go ahead and find that. Make sure you're on the correct graph as you do this. Please do practice it along with me because you will eventually have to do this on your own and you need to know how to go through the motions yourself. So it is one and a half as I guessed. So it's in between one and one and a half that both the velocity, the red graph, and the acceleration are both less than zero. The next question asks us to determine the speed of the particle when t equals one and a half. Now remember what speed is. Speed is going to be the absolute value of velocity. Remember our velocity function in this case was the quadratic that we had, the 6t squared minus 18t plus 12. So we want to evaluate that at one and a half. So I shall need you to go ahead and do that. When you do that, you should get negative one and a half. Please do try it and make sure. The speed though is the absolute value of that. So the speed in this case would be a positive one and a half. The units of measure of speed, remember if we did have units of measure would be the same as that of the velocity. The final question we're going to consider is to find the total distance traveled by this particle in between zero and four. And that's why I had us take a look at the graph of this first so we could get a sense of the path that particle was traveling because we want to determine the total distance traveled. So let's take a look at that again. And I'm going to go ahead and delete out of here the velocity and acceleration just so it's not quite as cluttered. So when we talk about finding the total distance traveled, really what we want is, I guess you can almost think the entire length of this curve if you want to think of it that way. So an easy way to do this and to organize your work is to consider certain critical points along the way. Well, obviously the particle starting at t equals zero. Remember what happens at the maximum in points of this curve. Remember that's where the derivative equals zero. The derivative of position is velocity. So we've talked about this before. When the velocity equals zero, that's where the particle is essentially stopping for a moment and then deciding if he's going to continue on forward or backwards. So right about here, it looks like about one is where the maximum is. So right there the particle stopping for a moment, then it continues on. It stops again here at two and then it continues on again after that. So what we want to do is figure out where it is the position of the particle at those key times at zero, one, two, and four, and basically add up the distances in between. So we can do that easily with a table and we're going to use our calculators to do that. So let's organize it by hand first and then we'll come back here and use the table feature in our calculator to help us determine this. So we're going to use the calculators table feature to find the position, the location, of the particle at each of these t values. So let's go back and create a table for that. If you had any x values there, you will want to delete them. So we have the t value zero, one, two, and four. And what you see generated are those y values are telling us the location of the particle at each of those time values. So it looks like at t equals zero the particle was down at negative four. By the time it got to t equals one, the particle was up at one. So in there it traveled a distance of five units. Going from one to two seconds, suppose it was, the position changes from one to zero. So in there we have another one unit that it traveled. Now we're not worried about whether the part particles going forward or backwards. We just want total distance traveled. All right, so in between t equals zero and t equals one it traveled five units total. Then it traveled another one unit in here. Then from two to four seconds it traveled 28 units. So if we add those up, that's how we'll get the total distance that the particle traveled. So in here it traveled five units, traveled another one unit in there, 28 units in here. So if you add those up you get 34 units. Is the distance total that the particle traveled?