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The velocity function (in meters per second) is given for a particle moving along a line. Find (a) the displacement and (b) the distance traveled by the particle during the given time interval.

$ v(t) = 3t - 5 $, $ 0 \le t \le 3 $

a) $-\frac{3}{2}m$b) $\frac{41}{6}$

05:52

Frank L.

02:19

Amy J.

01:47

Amrita B.

05:49

Linda H.

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Campbell University

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Okay, so let's have a look at this problem. Um we have a particle that's moving along the line and were given its velocity is three T -5. And we're interested in the displacement and the total distance traveled for that particle over the interval. Um where time is between zero and three. So the way that you would go about answering this question really depends on your level of expertise um in calculus so I'm going to explain it both ways. Kind of the most basic way and then the fancy calculus way. So um the first time we have to do is let's draw a graph of VT. Um so it's three t minus five which is just a fairly basic linear equation. Um and then we just want to remember any time velocity is negative, that means the particle is moving backwards, negative velocity. The negative part is just directional. So that means um from time equals zero which is right here at a time equals three. Um In this interval from zero until five thirds since the graph is below the y axis, that means the velocity is negative and that means the particles moving backwards. But then the particle switches um at time equals 5/3 and the particle starts moving forward because the velocity is positive. So it went backwards and then went forward. So it traveled um a lot of distance but it didn't get very far because it went backwards and then it retraced its steps if you will and then it went forward. So um let's see if we can figure out each question. So displacement. Um displacement is how far you wind up from where you started? Not how long did it or how much distance did you travel to get there? It's literally um if you go from point A. To point B. How far is 0.0.0.8 from point B. All right so um to find that answer we're going to find area the area under a velocity curve. Um Is the distance that it travels. So if we want displacement all of the time when the particle is going backwards which would be this would be the total distance would be um all of that area in the blue. And then to that we would want to add on all of the distance that it traveled going forward which is going to be all of this in the black. And so we can see that there's actually two triangles. So um the blue triangle. The area for that blue triangle area of a triangle is one half the base times the height. So base and then this is the height, the base of the triangle is 5/3. Um That's just the X intercept on this function. Um If you plug in zero for VF. T. Or for the Y. Value. Um then t would be able to five thirds. Um And then the height. The height is of the triangle. Is this distance here which is five is the actual height but we're going to actually count it as negative because the particles moving backwards. And so we want to show that it has backwards motion. So that's negative 26. Sorry 25 6. There we go. 25 6. Okay so I went backwards 25 6 ft or inches or whatever the units is. Okay then. Now let's look at the part where it's going forward. Um It has forward velocity um from um this point forward. So in the area to find that from time equals five thirds to time equals three it's moving forward. So to find the area that under the curve which is the distance that the particle travels in the positive direction we're going to find the area of that black triangle. So um base would be three minus five thirds. So that's going to be nine thirds minus five thirds which is four thirds. So the base is four thirds of the triangle and then the height is going to be four. So that is 16 60 total. Okay so if we want to find the displacement, what that tells us is for the displacement right? Um The particle traveled backwards so went negative 25 6th and then it went forwards 16 6. So it didn't travel as far forward as it had backwards. So we're going to expect a negative answer because we know that the particle never got back to its starting point since it went backwards. Um 25 6 and then only went forward 16 6. So um the difference between those is negative 96. which if you simplify that down is negative three halfs. So what that tells you is the particle wound up um to the left Because it's negative of its starting point. Um 3/2 which is 1.5 units from its starting point. All right so the displacement is 1.5 units left of the starting point. And if it started at the origin that would be um where the I think which I think it did say I think that would be where it started from. Okay so we'll just say the origin slash the starting point. Okay and then um distance traveled. A total distance traveled. So we've done most of the work to answer this. The total distance traveled is literally how much distance was actually covered in order in the whole entire trip. So the particle was trying to get from point A. To point B. Really should just walk from A. To B. Right? But instead particle went backwards a whole lot and then forwards a whole lot. So it traveled a larger distance. So if we want to find the total distance traveled I was called that tv t. Um we'll take it went 25 6 units backwards but I don't care about the direction it like you can think of it if it was walking 25 6 as how far it walked and then it walked more it walked 16 6. And the other direction. But here we're going to add those because we're trying to find the total distance that the particle actually covered. Um so when I add those together I get 41/6. Um and so that would be 41/6. Total units would be the total distance traveled. All right so that's kind of doing it using basic geometry. Um With a little bit of a calculus twist because of course you have to understand what we're doing in terms of calculus but of course there's much easier faster more efficient way of doing this. Um This is what an integral calculates a definite integral when you Find the integral from 0 to 3 three of V. Of T. D. T. That calculates the displacement. It calculates the area under the curve of velocity from 0 to 3. Um And just like what we just did with finding the area the triangles and the and science. The integral takes care of that for you. So in this case we're gonna integrate from 0 to 3. Um And then the function is three T minus five D. T. And um I think what I'll do is let me slide this down the way for down. No oh there we go. Sorry that took a little bit of work here. Okay so I will do it down here so you can see this little bit better. So from 0 to 3 and then our velocity function was three t minus five. And then of course we need to put in the D. T. We can put parentheses around that. So we need to find the anti derivative, the anti derivative of three T. So remember always bump the power up by one and then you divide by that new power. So we wind up with three halves T squared. Um and then minus and then the anti derivative of five would be five T. And we're integrating from 0 to 3. So um we are going to use the fundamental theorem of calculus. We are going to plug in the upper bound which is three into T. So I'll get everything ready copy that. We're going to plug in this upper bound of three. Right? And then we're going to plug in the lower bound of zero into three. Have to square in minus five T. Plug in zero. All right and we're gonna simplify this. So in this case we're going to wind up with 27 halves minus 15 minus and then all of this is just zero. So we have 27/2 -1515 is the same thing as 30/2. So we wind up with negative three hubs. Hey it's the same answer we got by just using triangles but we can see that we can use calculus in a rolls um definite integral. Will calculate the displacement of a particle when you take the integral of velocity. Okay, so then for total distance traveled that one gets a little bit more complicated when you go to use an inter role. What you're gonna want to do is you're gonna want to integrate velocity. But here's the thing we want to think of this as absolute value of velocity because remember velocity positive velocity means is moving forward when velocity is negative, it means the particles moving backwards. When we're interested in total distance traveled. We're we want all the distance whether it's forward or backwards, it doesn't matter. So we wanted to disregard the sign. We're gonna make it all positive. So if you go to graph what this would look like, right, it would be that same function from up above. But anywhere where the graph was negative, we would now flip it and it would become positive. So it would look like this. So essentially we would be finding this blue area but instead of it being below the X axis, is now above the X axis. And then we'll be finding this black area still. All right, so you can do this in a girl on a calculator. Um That's probably the most efficient way. If you're allowed a calculator for this problem. Um The other option is to use geometry which we essentially already did by doing the problem up here this way. Um or to break this up into two into girls. Um because this is uh the absolute value graph has a point that is not differentiable, so it's a little bit tricky to do it by hand. You have to actually use two separate intervals. Um when you do that, you'll get the same answer that we got before, which was 166. All right. Hopefully that helps to make sense and there is the answer to the question. Ok, meth.

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