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

Here we have an interesting problem. We're told a particle travels in a straight line, and its velocity is given on the graph as v of t. We're asked to find the displacement on the interval from zero to five. So how do we go about solving a problem like this? Well, in this problem, we're looking to find displacement. And we know that displacement is the integral from a to b of our velocity function v of t. But there's a problem here. Notice that we don't actually have a velocity function given to us as an equation. All we have is this curve on the graph. So what we're going to need to do is think about graphically how these integrals are represented on a graph and what we can go about doing to solve it based on what we can see. Now what I'm going to do is set up this integral a bit farther by putting in my bounds. I know my bounds are going to be associated with this interval that I'm given. So my lower bound is going to be zero, and my upper bound is going to be five. Now the function that we're integrating is the velocity function, v of t, and I'm integrating this with respect to t. Now to do an integral on the on a graph, recall that this is going to be the area that the curve of our function makes with the horizontal, or in this case, the t axis. And the way that I can calculate this, that I can find this area, is to just take a look at these portions of the graph that I have. Now my area is going to go from zero to five on the t axis. So we're going to start here at zero and we're going to finish over here at five. So to go ahead and find this well I'm going to be calculating this area right about there, which notice this part of the curve shows up above the t axis, and I'm also going to be finding this area down here which is the curve below the t axis. Now all of this area that's above the horizontal axis here that's going to be positive, all the area that dips below the axis is going to be negative. So once I calculate these areas here, I can just add them all up, and that will give me this integral. So first off, I'm going to deal with this portion of the graph before we dip below the t axis. Notice the triangle forms right here. And I know the area of a triangle is one half base times height. So if I wanna find the area of this triangle, well, that's just going to be one half of our base, which I can see our base goes from zero all the way over to three. And you can see our height is gonna go from zero on this vertical axis all the way up to three as well. So we have three right there. So we have one half of three times three, which all just comes out to 4.5. And that is the area of this portion for our graph. So now that we have our area there, we're going to find the area of this next portion of the graph, which dips below the t axis. So in this case, we'll call that area two. And area two is also gonna be one half base times height. Now the base, we start here at three, and then we go all the way over to six. We need to go over two places to or excuse me, go all the way over to five. When you go two places to get from three to five, you could also find the difference between five and three. So that means that our base here is going to be two since it's difference between five and three is two. Now next, we need to find our height. Now notice that our height starts here at zero on the vertical axis, but then we go all the way down to negative two. So because of that, our height is gonna be negative two. We need to go down two units to get that. And so to find our height, well, that's or to find our total area here, we can cancel that two with that one, giving us negative two as the area of this portion of the graph that dips below the t axis. So now have our area above and below the t axis, so we can use this to solve our problem here. So going back over to this integral, if I want to evaluate this, well, it's gonna be this area that we have, so I just need to add everything up. So it's going to be 4.5, this area above the graph, plus the area below the graph, negative two. Four point five minus two is just 2.5, and that's going to be our displacement and the solution to the problem. So this is how you can go about solving these problems, where you need to calculate a displacement based on a graph that's given if, say, we have a velocity graph, and we're looking to find our displacement based on what we can see. 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

A particle travels along the 𝑥-axis and its velocity is the given graph of $v (t)$ .
Find displacement on $open bracket 0 comma 5 close bracket$

2.5

4.5

-2

6.5

Verified step by step guidance

Step 1: Understand the problem. Displacement is the integral of velocity over a given time interval. In this case, we need to calculate the integral of v(t) over the interval [0,5]. The graph of v(t) is provided, and it is a straight line decreasing from 5 at t=0 to -4 at t=5.

Step 2: Analyze the graph. The velocity graph forms a triangle above the x-axis from t=0 to t=2 and another triangle below the x-axis from t=2 to t=5. The areas of these triangles represent the contributions to displacement, with positive area above the x-axis and negative area below the x-axis.

Step 3: Calculate the area of the triangle above the x-axis (from t=0 to t=2). The base of the triangle is 2 units (from t=0 to t=2), and the height is 5 units (from v=5 to v=0). Use the formula for the area of a triangle: A = (1/2) * base * height.

Step 4: Calculate the area of the triangle below the x-axis (from t=2 to t=5). The base of the triangle is 3 units (from t=2 to t=5), and the height is 4 units (from v=0 to v=-4). Use the formula for the area of a triangle: A = (1/2) * base * height. This area will be negative because it is below the x-axis.

Step 5: Add the areas from Step 3 and Step 4 to find the total displacement. Remember to account for the sign of the area below the x-axis. The sum of these areas gives the total displacement over the interval [0,5].