Velocity, Acceleration and Distance 1,747 views
Now that it?s time to discuss velocity and acceleration, we?ve better speed things up a little bit. Speed things up, accelerate. Yeah that was a joke, you probably didn?t get that. Anyhow, velocity, distance, acceleration are some of the most important things involving Calculus. At least I think so, because I?m a physics guy. Remember that a lot of Calculus at least for the start was developed by Isaac Newton in order to make his Physics work. Besides the great Isaac Newton, you also need to know this stuff, because there?s a lot of it on the AP test.
Velocity; it?s a rate of change. It?s the rate at which distance is changing with time. Let?s have a look. Here is your old friend; distance equals rate times time. You got that one around about fourth grade maybe. If you take this and solve it algebraically so that rate is by itself, you?ve got rate that equals distance over time. By distance what you mean is how far the location changes. It?s often called change in distance. Then the time is really the change in time, how much time went by. Well, these are differences. If you make differences smaller and smaller and smaller just like you do all the time in Calculus, instead of having delta d, you have dd, as in dd over dt.
Yeah it?s a derivative. All you have to do is take a distance equation; take its derivative with respect to time and you?ve got a velocity equation. Take that one step further and acceleration is the rate of change of velocity over time. So acceleration is the derivative of velocity. Velocity is the derivative of distance, and it reverses as well.
Here is the basic kinematic equation. You almost certainly had this in your Physics class. You might not have had the form in it that has d0 in it. D0 is just the starting location of an object. This is pretty cool. I love showing this. Remember what we just said; velocity is the rate of change of distance. So I?m going do d over dt to both sides to find the derivative. So I have d over dt on both sides of d of t. Don?t let it confuse you that you get d here and d here. This is the differential. This is d for distance. I know it?s confusing. Sometimes to avoid this confusion, you will see the distance listed as r. It?s pretty common.
We have to do the derivative of both sides. We?ve got ½ 80²+ v0t+d0. Derivative of this one is the velocity with respect to 10. Let?s see, we have to do the derivative. The variable here is t. That?s one thing that can confuse you. A for this equation is a constant. It?s just a number; v sub 0 that?s the starting velocity. That?s just a number. D sub 0, that?s the starting location. That?s just a number. It makes a difference when you do the derivatives. So the only working variable here is actually the t. Derivative of t² is 2t times the ½ just gives you 80. The derivative of this it?s really just one power of t, one power less is 0 powers. All that?s left of that is v0. The derivative of that constant is just 0, look at that. It?s another one of the Physics equations.
Velocity equals acceleration times time, plus the starting velocity if you had one. Taking it a step further, if I do this derivative, I?ll have the acceleration. The derivative of acceleration times time, time being the only variable here is just acceleration. That?s an unchanging velocity. It?s a constant, so its derivative is 0.
Time for a little practice. What we?re going to do now is use derivatives, velocity, and acceleration together. Let?s start and see what we?re given. We are given distance. A particle moves such that its distance from its starting point is given by the equation d of t equals t² times e to the 3t. Find its velocity. It?s really important to look and see which one you?re given, what you?re supposed to find so you know what order to go.
Now we just did something involving derivatives. This one is also going to involve a derivative, but you need to be prepared that it can go the other way. The derivative of distance is velocity, but you may have a question coming up where you?re given a velocity equation and you?re asked to find a distance. Reversing it involves an integral.
Let?s go to work. We?re supposed to do the derivative of this. So the velocity at any given time is the derivative of the distance d prime of t. So t² times e to the 3t, a variable factor. This one is going to take the Product Rule. Remember the Product Rule is the derivative of the first part times the non-derivative of the second plus the non derivative of the first factor times the derivative of the second factor.
Derivative of the first factor is 2t, times the non-derivative of the second factor, plus the non-derivative of the first factor, times derivative of the second factor. That?s not just e to the t. It?s e to the 3t. This is a Chain-Product combination. Well, if I do the Chain part of this, the derivative of 3t is just 3. The derivative of e to the 3t, the remaining part is e to the 3t.
We?ve got an equation, we can simplify this. Let?s group things up a little bit. Notice they both have an e to the 3t factor. Let?s take that up. In the long run that actually simplifies things a little bit for you. So we have 2 times t plus t² and we factored out e to the 3t power. There we go. This little bit is done. We?ve got that velocity. Are we done? We?re not.
You may have some problems where finding the equation is equation. This is not one of those. This is one where you?re supposed to find the equation and actually use it to do something. Put time equals 3 in there. So we have v of 3 equals 2 times 3 plus 3² times e to the 3 times 3 for time again.
If you work this one out, you will get an approximate velocity. This one is big. It?s 267,000. One last thing to take look at is check the units. This problem didn?t give us meters or miles or anything for distance. It didn?t give us seconds, or hours or any time unit for time. So we don?t have to have any units in the answer. This is just wanting to prove, but that?s the way it is. For this one you don?t need any units.
The next problem we?re going to do is going to take it a step further; finding the acceleration. To save a little bit of time, I set this up so it?s a continuation of the last problem. When you see this on a test, they may not have continuations. You might just be told to start with this. In fact you might even be told to start with distance and then take both steps to find the acceleration before you can even substitute.
This time though, we?ve already worked out what the velocity equation is by taking the first derivative in that last equation. This is the equation; this is the factored form that we did. Well, acceleration is how fast velocity is changing. How fast something is changing, is effect of their rate. We?ll find in the rate that the velocity changes with respect to time. Rates with respect to time are derivatives with respect to time.
Like I showed here, you?re going to find the acceleration. It?s the derivative of the velocity. This is the velocity equation. We have to do a d over dt at the derivative of that part. One of the reasons that I factored this out was also because, believe it or not, it makes this derivative easier to do. So we have Product Rule here. The first factor is 2t plus 3t². I?ll take its derivative. Its derivative would be 2 plus 6t, times the non derivative of the second part, e to the 3t, plus the non-derivative of the first part. That is 2t plus 3t², times the derivative of the second part. That second part derivative takes the Chain Rule again. It?s going to be 3e to the 3t.
3e to the 3t; it is a little harder to see there. Well, we?re almost done. We were asked to find the acceleration at a particular time. We were told to find the acceleration at a time of 3. So I?m going to leave this for you to try out if you want some practice, but basically you just put 3s in for all the ts. In this case the acceleration works out to be 462.
One thing that might be confusing you especially if you?ve had a lot of stuff in Physics, you might be thinking wait a minute, isn?t acceleration always 9.8? No, that?s only the acceleration caused by gravity and only on earth surface actually. It is possible to have accelerations that are not constants like that. You can have different accelerations. This is a non-constant acceleration that varies with time. It can happen. The reason it?s not familiar to you is because High School Physics rarely goes that far.
So where do integrals fit in? Well, if velocity is the derivative of distance, then the integral of velocity has to be the accumulated distance. They have to be able to reverse each other. Here is a common kind of problem. The problem says that a particle moves along the x axis so that anytime where time is greater than 0, the velocity is given buy this equation. Another bit of information you?re given is that the particle is positioned at x equals -100 at time equals 3. That?s asked me to find the object?s position at time equals 9. Something that can confuse you on this is that there?s three numbers floating around 3 times. You have to decide which ones are going to be your limits of integration. There?s the 0, there?s the 3, and there is the 9.
The one we?re going to ignore is the 0. The reason we?re ignoring the 0 is because we have a starting position at time equals 3. We just have to find out how much distance got accumulated between 3 and 9 seconds. If you look at this graph, this graph shows time versus velocity. Velocity starts at high, but it?s positive. So the object is moving forward. It keeps on moving forwards yeah, I know it looks like its moving forward doesn?t it? But really this is just velocity. This is a positive velocity, so at this time it?s moving forward. Positive velocity moving forwards, positive velocity moving forwards.
Down here though, it?s got a negative velocity. It?s moving backwards, but we don?t need to worry about that. The integral will take care of the moving backwards. If you started at distance of -100, maybe you?re chugging along and get to +500 before the velocity is backwards and knocks you backwards to maybe +450. We don?t have to worry about it. We are going to do an integral though. The integral is going to be between 3 seconds and 9 seconds. So looking at the graph, it seems like we accumulate some positive distance forwards, and then lose some of it going backwards.
Let?s set up our integral. We need to find the accumulated distance. That is accumulated between times 3 and 9. So the integral is between 3 and 9. You integrate velocity to get distance. The velocity equation is 200 minus 7t ln of 10t, dt of course, because that?s the working variable of time.
Well, at this point you look at this thing and say oh men how am I going to integrate that? You don?t have to worry; you wouldn?t likely see kind of an equation on the part of the test where you?re not allowed a calculator. Yeah it?s possible to integrate it, but by the time you substitute the 3s and 9s in, you need decimal approximations anyhow. So they wouldn?t give you this on the non-calculator portion.
All you have to do is set it up in your calculator. Let?s write it out on more time. Set it up in your calculator so that your accumulated distance is the integral from 3 to 9 of 200 minus 7t ln 10t. Type this portion in to have your calculator graph it, and then use the integrate function to find the area between 3 and 9. In this case, your calculator is going to tell you that the accumulated distance is 157.441. That?s the accumulated distance. Look over the problem for a second. I?m going to pause for a moment while you try to figure out the last thing we have to do. It?s a bit of information we haven?t used yet. The hint is this is accumulated distance.
We started at -100. So we started 100 back. Put this on a nice little number line. There is 0. We started 100 back and the particle moved forward away and then it moved backwards away and wound up at a finishing location. That?s what we?re supposed to find. It?s the finishing location. Basically all you have to do is take the starting location of -100 and add the accumulated distance to that. So +100 gives us a final position of 57.441. I?ll leave off the 441, because I?m out of screen.
The most difficult kind of problem you?ll be asked to do are ones that involve distance, not position, but distance. Let me explain the difference. Let?s say I start off here at a location of 0 and move forward 2 meters and move backwards 2 meters. If I made a velocity equation for that, and integrated it across the time I was moving, it would have told me that the result was 0. I didn?t go anywhere; I wound up at the same place that I started. Sure my position at the end is where I started, but the distance I travelled was 2 forwards and 2 backwards. Distance doesn?t care about direction. I travelled 4 meters. Let?s try it.
This problem says a particle moves along the x axis so that anytime its velocity is given by 1/3t³ minus 4.The particle is in position 5 at time t equals 0. Don?t look for the 5 on here, because this is not a graph of position. This is a graph of velocity versus time. So we know at time is 0, position is 5. We?re asked to find the total distance travelled between the time of 0 and the time of 3. If I put 0 on the graph, or substitute into the formula, you?ll find out that the velocity at that instant in time was -4. Moving backwards, we?re going to have to account for that. At time equals 3, the velocity is going to be positive. If I just did a single integral all the way across here, the integral would take this area and make it negative. And then add that negative to this positive area to get a result that would be too small.
What I?d be finding is the location. I want the distance. The major steps we?re going to take on this are to look at the graph, determine which regions were moving positive and negative in. This is a region with negative velocity. This is a region with positive velocity, and then set up two separate integrals. We don?t use the 5 at all. It?s just a spoiler number. If I start here and move a meter forwards, I?ve gone a meter. If I start here and move a meter forwards, I?ve still gone a meter. How far I went doesn?t depend on where I started. So let?s do the set up. We?ve got to find out where that is so we have our limits of integration.
Now depending on where you are on the test, you might be able to just type this into your calculator as a graph. And tell the calculator to use the 0 function which finds the 0s of the equation. I set this one up so it could be done manually, because you might have to do these on the non-calculator portion of the test. You find 0s of course by finding when the up and down coordinate in this case v of t equals 0. So let?s substitute 0 in there. If you are given one of these to do by hand, it?s not well likely that they?re going to be on your really complicated Algebra; maybe a little harder than this, but not too much.
We?ll solve this little guy. I?ll add 4 to both sides, so we get 4 equals 1/2t² and we will multiply by 2 so we?ll get 8 equals t². That looks like that?s going to gives us t². I lost a variable. This is t³. I?m sorry about that. t³. So what I get is a value of t equals 2; nice and neat.
But if you?re asked to do this by hand, it?s going to be a fairly nice and neat number which doesn?t include fractions by the way. Fractions are nice and neat. That?s 2. So I need to do an integral between 0 and 2. I need to do an integral between 2 and 3. When I set it up, I?m going to make this integral negative, because the integral itself will give me a negative result. I have to do the negative of that in order to get the distance that?s travelled.
Here again are the steps. We?re about half way through the problem now. We found the 0s. Now we?ve got to test to find the positive and negative regions, because again you might not have the graph. If you?re on the non-calculator portion, you won?t have that pretty picture that we just looked at. You need to do manual tests. Then you set up your integrals, and integrate.
Here is what we know. On the last part of the problem, we found out that we have a situation like this. Time equals 0 is one of the end points of our interval. The place where it changed from negative to positive, we don?t know yet is at 2. The reason I say we don?t know that yet is because I?m pretending now that I haven?t seen that graph. I don?t know the velocity was negative. I?m doing the old fashioned manual test like I did in Pre-Calculus to find out where the function is positive or negative in a particular region.
In points of the region, I'm going to call this region 1, and I?ll call this region 2. I just need to do a simple sign test to see if the times between 0 and 2 give me a positive or a negative result. I don?t even care what the result is. I just care if it?s positive or negative. Pick anything you want between 0 and 2. And since I can?t pick anything, I?m going to make my life simpler and pick one. Why not? If you pick anything in the region, you?re going to get a positive or negative result. If I pick this one, I get negative or if I pick in this, I?d still get negative. If I picked that, I?d still get negative.
Anywhere in the region you?ll get the same sign on the result. So our first test I?m testing t equals 1 and I?m going to find v of 1. That?s easy enough to do. Let?s make that one a 1 not a t. There we go; v of 1 equals ½ times 1 to third power minus 4. That?s a half of 1 minus 4. It?s negative. That?s all we cared about. That is negative.
So I know when I set up my integral, I?m going to have to do the negative of that result. Next region might be positive or might not be. You can probably predict it, but don?t assume that because this was negative, the next will be positive. You never know. Between 2 and 3 is 2.5 if that is anything and now we?ll find the velocity at 2.5. That?s ½ of 2.5³ minus 4. Now you might panic and think ½ times 2.5³ minus 4, how am I going to do that without Mr. Calculator? Well, you could write it out by hand, but if you recognize a little fact about this, you can save yourself a bunch of time.
Remember that 2³ is 8. ½ of 8 is 4. 4 minus 4 is 0, so that?s on the border line. Anything more than 2 if you cube more than 2, you?re going to get more than 8. Half of more than 8 is more than 4. More than 4 minus 4, it?s positive. So we know this is a positive region. Let?s go set up our integrals we?re close to done.
We just found out that we have a negative velocity between 0 and 2. Well, I?m accumulating my distance now. I have to do an integral from 0 to 2 of that equation. So that?s ½t³ minus 4dt. Because anything between 0 and 2 produces a negative velocity, this integral is going to give us a negative distance. We don?t care about the direction when we?re talking about the distance travelled. So put a minus in front of that. That?s the first section.
The next section has a positive velocity, so I?ll just add in its integral. I?ve got the integral from 2 to 3same thing again; ½ of t³ minus 4dt. Now let?s do the integral. Don?t lose this negative. It?s really easy to do. If you do, the problem will fall apart.
It?s very important to learn these kinds of things to lock those positives and negatives really closely. Integral of t³ is ¼ t to the fourth, remember it?s a correction factor, times this ½ that was already there. So I?m going to have 1/8 t to the fourth. Integral of -4 is -4t. You?re going to integrate from 0 to 2 plus, the nice thing here is that it?s the same integral. So you don?t need to do those steps again. Just copy this down with new limits of integration. New limits of integration on the same thing gives us this; from 2 to 3.
Now be very careful when you substitute. Don?t forget this negative, and really take your time to get the positives and negatives correct. Or again you?ll get the wrong answer. I?m going to save a little bit of time though. That?s a whole lot of fractions to write out. The result of this one is the distance that you accumulate is 81 over 8. Well, again what about that starting position? The 5 doesn?t matter. We just worry about how far we went, not where we started. So you don?t use the 5.
Well, I?ve just given you a background on one of the really big topics in Calculus; distance, velocity, acceleration and how they?re related. You can do derivatives, or you can do integrals depending on what information you?re given. If you are given distance, you can do the derivatives to get velocity, and then another derivative to get acceleration.
If you?re given an acceleration equation, you can do the integral of that to find a velocity equation. You can do the integral of that velocity equation to find the distance equation. Great stuff, it?s lots of fun. It?s incredible useful for Calculus. It?s incredibly useful for Physics. If you go on eventually to take some college Physics, you?ll probably take Calculus based Physics where you?ll see this stuff in more detail. Well, be sure to check out the bonus materials. They?re some good practice problems there involving all of the steps all written out so you can check your work.