Guys, in this problem, we have a remote control car that is subject to a force that varies over some distance here. So, when we have these force-distance graphs, we're going to take the area under the curve, and that's going to be the work. We want to calculate the speed of the car at \(x = 4\). Now, the car initially starts from rest, so this is \(v_0 = 0\). We want to calculate the final speed. So, how do we do that? We're going to relate the work that's done to the change in kinetic energy by using the work-energy theorem. The net work is equal to the change in kinetic energy. I'm going to take the area under the curve for this whole entire graph here up until \(x = 4\), and then I'm going to relate that to \(K_{\text{final}} - K_{\text{initial}}\). I'm going to break this up into a couple of sections. The work done from 0 to 2, plus the work done from 2 to 3, plus the work done from 3 to 4. That is going to be equal to \(K_{\text{final}} - K_{\text{initial}}\). However, because the initial speed is equal to 0, there is no initial kinetic energy and all of this is going to be final kinetic energy, which then I can relate to the speed of the car. I'm going to take a look at the areas under the curves for each one of the three little terms that I've made. The work done from 0 to 2 is basically going to be all of this area right here. I'm going to highlight this in blue. That's going to be the sky. What about from 2 to 3? Well, from 2 to 3, there actually is no area under the curve because we're at 0. There is no force acting on this car from 2 to 3, so there is going to be no work done. Finally, from 3 to 4, there's definitely going to be some work done because it's going to be this area here, under the graph. I've got those highlighted areas here. Let's go ahead and start calculating.
So, if I'm going to calculate the work done from 0 to 2, I'm basically going to cut this up into 2 shapes: one triangle and one rectangle. I'm going to do this all in one sort of line, but this is going to come down to 2 terms. I've got the area of a triangle which is \( \frac{1}{2} \times \text{base} \times \text{height} \), which is going to be \( \frac{1}{2} \times 1 \times 20 \), and then plus the area over here, which is really just a rectangle. This is going to be base times height. So the base is 1 and the height is 20. So, if you go ahead and work this out, what you're going to get is you're going to get 30 joules. So, 30 joules is the first little section of this right here. So, actually, let me highlight this in blue. Let's go ahead and calculate the work done from 3 to 4. The work done from 3 to 4 is just going to be this little triangle right here. I've got \( \frac{1}{2} \times \text{base} \times \text{height} \), that's the base. And then the height of this thing, well, this doesn't hit all the way down to -20, and it's a little bit past -10. So, what I'm going to do is I'm going to call this -15 here. It's right in between. And remember, it's negative because we're in the negative y-axis here. So, the work that's done is actually going to be -7.5 joules, and that is the work done from 3 to 4. So, now, I really just add these two things together. And what this just becomes is I have 30 joules plus, this is -7.5 joules. And this equals \( \frac{1}{2} \times m \), so we know what the mass is. This is actually going to be 4 times \(v_{\text{final}}^2 \) here. So, this is \(v_{\text{final}}^2 \). I get 22.5 joules equals, this is going to be 2 \(v_{\text{final}}^2 \). So, you go ahead and work this out. What you're going to get is you're going to get 3.35 meters per second. So, that's the final answer here. That's the speed of the car at \(x = 4\). Let me know if you guys have any questions.