A particle travels along the 𝑥-axis and its velocity is the given graph of v ( t ) v\left(t\right) v ( t ) . Find total distance on [ 0 , 5 ] \left\lbrack0,5\right\rbrack [ 0 , 5 ]

Here we're going to continue on this problem that we had before where we had a particle that traveled in a straight line, and we were given the velocity as a function v of t on a graph. Now what we need to do in this problem, for this portion, is find the distance based on this graph from zero to five, the given interval. Now in the previous problem, we calculated displacement. And recall that we found displacement by integrating our velocity function from a to b. And that's gonna be the velocity function integrated with respect to t. Now in this problem we're not dealing with displacement instead we are dealing with distance, and the difference between distance and displacement is that distance cannot be negative. So what we need to do is focus on the absolute value of our velocity function and that's going to allow us to solve the problem. Now how exactly do we integrate an absolute value function, especially when we're dealing with a graph? Well, all we really need to do for this problem is focus on the area that's under the curve of this function, but we can't actually dip below the graph at any point. So any area that shows up below the t axis, we're going to need to reflect that curve upwards so we can go ahead and calculate that area. Now I'm gonna go ahead and write this integral a bit better so we can calculate our distance. Writing this, we're going to write the integral based on the interval given to us from zero to five. That's going to be the absolute value of our function v of t. That's the velocity function. So So we're integrating this all with respect to t. And to go ahead and do this integral, well, what I recognize based on the graph is we're going to find this area right here. That's the area that shows up above the t axis. Now we have another portion of this graph that goes below the t axis, axis, but I can't focus on this portion since it goes below it. So what I have to do is take this portion of the graph and reflect it upwards like this. And then this is the area that I can find right here. So we're going to find this area that all shows up above the t axis now. So everything that we calculate should be positive, hence the absolute value that we had on our velocity function. Now keep in mind this interval, just like we had before, is going from zero to five. So we start here at zero on our graph, then we go all the way over here to five. Now from here, I can go ahead and calculate my area. And I can already see that they're going to be triangles that show up in this problem. And I know the area of a triangle is one half base times height. So first off, I'm gonna find the area of this first portion here. The area is gonna be one half of the base, which I can see the base goes from zero all the way to three, then we'll have that multiplied by the height, which goes from zero all the way to three on the vertical axis. So we have one half three times three, which is 4.5. And that's going to be our area right here. Now next, what I need to do is find my next area. And what I'm gonna go ahead and do is label this whole thing as area one since that's the first portion there. Then we'll find an area two as well, which is gonna be this next portion of the graph. Notice another triangle forms so I can use the same equation, one half base times height. Now first off, I need to find the base of this triangle. Notice we go from three all the way over to five. It takes two places to get from three to five. The difference between five and three is two, so two is gonna be our base. Now next, I need to find the height. Well, starting on this vertical axis here, we start at zero, and then we have to go up two units. So that means that our height is going to be two. I can cancel the one half with this two, leaving me with just two as the second portion of this graph for the area. So now that I've found this, calculating the distance should be pretty straightforward. I just need to find this total area, which I can go ahead and add up the results that I got, which is 4.5 plus two. Four point five plus two comes out to 6.5. That is going to be the area underneath the curve of our velocity function when we reflect this portion of the graph up. And that is going to be the solution to this problem. So this is how you can solve these types of problems when you're dealing with these graphs, and you need to figure out what the distance is on that graph. So hopefully this gives you an understanding of how to find distance and displacement on a graph if you are given some velocity function v of t. So hope you found this video helpful, and let's move on.

10. Physics Applications of Integrals

Kinematics

Calculus

10. Physics Applications of Integrals

Kinematics

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Multiple Choice

A particle travels along the 𝑥-axis and its velocity is the given graph of $v\left(t\right)$ .
Find total distance on $\left\lbrack0,5\right\rbrack$

2.5

4.5

6.5

Verified step by step guidance

Step 1: Understand the problem. The graph provided represents the velocity function v(t) of a particle moving along the x-axis. To find the total distance traveled on the interval [0,5], we need to calculate the integral of the absolute value of v(t) over this interval.

Step 2: Analyze the graph. The velocity function v(t) is a straight line decreasing from 5 at t=0 to -4 at t=5. This indicates that the particle changes direction at t=2, where v(t) crosses the x-axis (v(t)=0). The total distance traveled will be the sum of the absolute values of the areas under the curve from t=0 to t=2 and from t=2 to t=5.

Step 3: Break the integral into two parts. Since the velocity changes sign at t=2, split the integral into two intervals: [0,2] and [2,5]. For the interval [0,2], v(t) is positive, so the integral is ∫[0,2]v(t)dt. For the interval [2,5], v(t) is negative, so the integral is ∫[2,5]|v(t)|dt, which is equivalent to -∫[2,5]v(t)dt.

Step 4: Determine the equation of v(t). The graph is a straight line, so v(t) can be expressed as a linear function. Using the points (0,5) and (5,-4), calculate the slope m = (y2-y1)/(x2-x1) = (-4-5)/(5-0) = -9/5. Thus, v(t) = -9/5*t + 5.

Step 5: Set up the integrals. Substitute v(t) = -9/5*t + 5 into the integrals. For the interval [0,2], calculate ∫[0,2](-9/5*t + 5)dt. For the interval [2,5], calculate -∫[2,5](-9/5*t + 5)dt. Add the absolute values of these two results to find the total distance traveled.