Suppose that a particle travels along the x x x -axis and its velocity is given by v ( t ) = 2 cos t v\left(t\right)=2\cos t v ( t ) = 2 cos t for 0 ≤ t ≤ 2 π 0\le t\le2\pi 0 ≤ t ≤ 2 π . Find total distance traveled by the particle on [ 0 , 5 π 4 ] \left\lbrack0,\frac{5\pi}{4}\right\rbrack [ 0 , 4 5 π ] .

So in this problem we're told suppose that a particle travels along the x axis and its velocity is given by this function right here two cosine of t, and we're also given the time interval from zero to two pi, and asked to find the total distance travelled by the particle on the interval zero to five pi over four. Now how can we go about solving this problem? Well, notice that we're asked for a distance traveled. And when we're trying to find a distance, what we need to do is integrate our velocity function, v of t. And we're integrating this from the initial time to the final time, but remember distance is a value that cannot be negative. We can't get a negative result. So we're looking for the absolute value of our velocity function that we're integrating with respect to t. So in this problem specifically we're going to be integrating on the interval that's provided for us zero to five pi over four, and then we're going to have the absolute value of the velocity function. Well the velocity function is given right here as two cosine t. So So we're gonna have two cosine of t, and we're integrating this whole thing with respect to t. Now here's where things get a little bit weird. Notice we've got a couple situations that make this look a bit complicated. We have to deal with these absolute value bars, and we also have to deal with this cosine term. But there are a couple things I can do to make things simpler. Because this two here is just gonna be a constant here. So it's gonna be multiplied by the whole thing anyway, not impacting whether or not we get a positive or negative result. So I can take that two and pull it outside of the integral here. So we won't have two, and that's going to be the integral from zero to five pi over four, and it's going to be the absolute value of the cosine of t. So this is what we get. And to find the integral for the absolute value of the cosine of t, we need to think about what the cosine of t graph looks like. This integral would give us the area underneath the curve, but how exactly would all of this work? Well, imagine that we have some kind of curve. So go ahead and draw a y axis here, and then we'll draw our t axis right here. Now we know the cosine looks something like this, where we get this kind of curve and oscillating pattern, and this is just gonna keep on going but I'll just stop it right about there. Now we're trying to find the area under this curve for the cosine of t. Now if I go ahead and look at this first portion here, we would have this area right here, and that would be part of the the area that we're trying to find. But we also have all this area that goes below the graph right about there. This area would be calculated as negative, and we can't get a negative result. So what we would actually need to do is take our curve and reflect it over the graph like this and then make this area positive. So we're, in essence, just focusing on this portion of the graph above the above the the place where the y values will be negative. We're focusing on this area right here, and we're ignoring any area that dips below this graph. So we're not worrying about that. Now doing this is gonna be a bit tricky, but what we can focus on is this portion right here. This portion goes from zero all the way over here to pi over two. After that, you're gonna keep on going this way. Now as you keep going this way, five pi over four is somewhere around right here. So about right there would be where we land on the t axis for five pi over four. Meaning that the area that we're gonna focus on is all of this area right here. So we need to figure out how we can find this area, and to do that well after looking at a graph I think I know how to approach this problem from here. So first off, what I'm going to do is I'm gonna take this and split it into two separate integrals where we're only focusing on the area that's positive. And the way that I can do that is by taking this two on the outside here, and then first off, we'll have this integral that goes from zero, and then we're gonna go all the way up to pi over two because that's this portion that we know is already above the axis. We're above the t axis here, so we're going to have the integral of the cosine of t with respect to t. But now what we have to do is add on the next portion of the integral, and the next portion of the integral is going to go from pi over two, where we left off right here, all the way up to five pi over four. But notice that these values would typically dip below the t axis. The reason that we reflected it up is so we get positive values. So I need to change the sign on this entire integral here since that's going to be all these values that will be above the t axis here. So because of that, we'll have the integral of this cosine of t, and then that's going to be with respect to t. So notice how we've gone ahead and taken this whole thing, and we've only focused now on the positive values, which means we can just evaluate these integrals separately. So we're gonna have this two on the outside here. And to go ahead and do this, well, first off, deal with this integral, the integral of cosine of t, which I know that the integral of cosine is just gonna be sine. So we're going to have the integral, which is gonna be sine of t. This is gonna be bounded from zero to pi over two plus, and then we're going to have the integral here, which is also gonna be sine. So we're going to have plus the negative, and then that's going to be sine that we have right here. So we're gonna have the sine of t. And then this whole thing is going to be bounded from pi over two to five pi over four. So this is what we get. Now from here, we're gonna keep going. So we'll have this two on the outside here, and what I can do is just plug in my pi over two since I know zero is just gonna set the whole thing to zero. The sine of zero is just zero, so we're gonna have the sine of pi over two as the only place that we plug in this upper bound. Then this whole plus negative here, I'm actually just gonna change that to a minus, then we're going to go ahead and deal with this portion of the integral or this portion of the math that we need to work on. So what's gonna be here is we're going to have the sine of our upper bounds, we're going to have the sine of five pi over four, then we're going to have minus and that's going to be the sine of pi over two, and then this is going to be everything we have within these brackets here, and that's going to be what happens when we plug all of our bounds in. Now from here, encourage you to use your calculator to evaluate all of this, and you wanna make sure that you are in radian mode when dealing with this since we can see that we have pis and pi over twos that we're dealing with. So if you go ahead and plug this all in, don't forget to multiply this two on the outside, you should get approximately 5.41. So that's gonna be your result, and this is going to be the total distance traveled by the particle. So that right there would be the solution to our problem, and that is how you can find the distance. So as you can see, it did take a bit of work to get there, But the main thing that we needed to focus on is dealing with this integral, just making sure that we only focused on the values that would come out to positive as an output, which is what we did. So that's why we focused on this place here on the graph, which is where the y values are positive above the t axis, and that allowed us to find this result of 5.41. So hope you found this video helpful. Let me know if you have any questions, and let's move on.

10. Physics Applications of Integrals

Kinematics

Calculus

10. Physics Applications of Integrals

Kinematics

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Multiple Choice

Suppose that a particle travels along the $x$ -axis and its velocity is given by $v\left(t\right)=2\cos t$ for $0\le t\le2\pi$ .
Find total distance traveled by the particle on $\left\lbrack0,\frac{5\pi}{4}\right\rbrack$ .

1.41

2.00

5.41

0.027

Verified step by step guidance

Step 1: Understand the problem. The velocity function v(t) = 2cos(t) is given, and we need to find the total distance traveled by the particle over the interval [0, 5π/4]. Total distance is calculated by integrating the absolute value of the velocity function over the given interval.

Step 2: Set up the integral for total distance. The formula for total distance is: ∫|v(t)| dt over the interval [0, 5π/4]. Since v(t) = 2cos(t), we need to evaluate ∫|2cos(t)| dt from t = 0 to t = 5π/4.

Step 3: Determine where the velocity changes sign within the interval [0, 5π/4]. The velocity changes sign when cos(t) = 0. Solve cos(t) = 0 to find the critical points within the interval. These points will help us split the integral into subintervals where the velocity is either positive or negative.

Step 4: Split the integral into subintervals based on the critical points found in Step 3. For each subinterval, evaluate the integral of 2|cos(t)| dt. Remember to take the absolute value of cos(t) to ensure the total distance is calculated correctly.

Step 5: Add the results of the integrals from each subinterval to find the total distance traveled by the particle over [0, 5π/4]. This sum represents the total distance traveled, as the absolute value ensures all contributions are positive.