In this problem. We have two cars and B which are initially next to each other at the origin at rest and they begin accelerating with different accelerations and stop at different times and velocities. And we want to know what the acceleration versus time, velocity versus time and the distance versus time graphs look like so first for the acceleration versus time graph. This is most straightforward for a it's initially at 4 m, so for meters per second square and it's like that for 10 seconds and then it maintains a constant velocity. So the acceleration is then zero for be. It's at 5 m per second, squared 5 m per second squared and it does this for some amount of time, which will figure out right now. So it is accelerating at five million stuffing sweat and has a final velocity of 25 m per second. So we know that the is equal to be not plus acceleration times time, the initial velocity of zero, so we can just plug in our final velocity of 25 m per second, which is equal to our acceleration of 5 m per second times the time and we find that the time is just 25/5, which is just five seconds. So this should be a half as long as a is accelerating for so that's five seconds. And then again, it's just zero because it is maintain a cost in velocity. Next, we want to know the, uh, with the velocity grabs look like So we know what the slope of thes are going to be initially because the slopes are going to just be equal to the acceleration. And then once they're done accelerating, they're going to be a flat line since they're maintaining a constant velocity. Will do be first, since we already know what the final velocity is going to be. So for be we're gonna have some line. And this line is going to be such that the final velocity is 25 m per second and the time at which this occurs is five seconds. And then after that, it's going to be a straight line. So that's what B will look like. And for a Why don't we go ahead and find the final velocity using this equation so we know the acceleration of the time. The initial velocity of zero. So like we did last time vehicles 80 which is just four times 10 which is it's time, which is just 40 meters per second. So a it's going to have a more shallow uh, it's gonna have a less steep line. Can be because the acceleration is lower. So it's going to look something like that if we can draw a straight line and these air both should both be starting at deep oops at the origin and then or pretend this is the access. But this is going to be 40 m per second and the time which again pretends on the access, is going to be at 10 seconds. And then after that, it's going to be a straight line. No, the last for this problem is what does the position look like over time. So the what broadly is goingto happen? Is there going thio? They're accelerating initially, so they have some like parabolic shape that and then they have a constant velocity. So then they will just be a straight line so blue, since it has a steeper acceleration, is going toe look like this and then it's gonna have some constant velocity that it's gonna go off into space and red is going to have a more shallow Rapolas. Since the acceleration is less, it's going to reach a gonna be going for longer, and then it's going to have a higher speed than so it's going to be a straight line after that. That's going to be steeper than blue, since the final velocity is greater. And if we say this is around where it stops or I guess maybe this is around where it stops decent accelerating. So this is 10 seconds where it stops accelerating. And for Blue, where it stops accelerating is five seconds, and we want to know. In addition to this, what the distance between the cars is when T is equal to 15 so T equals 15 so that's somewhere over here. So what we can do for this is we can use the equation that the change in the distance is going to be the initial velocity times time plus one half of the acceleration Times T squared for an Excel constant acceleration. And then after that, the, uh, change in distance. It's just going to be the velocity times the time. So we can add those up, then Thio get the total distance changed for both of them. So we'll start with a so delta us. The initial velocity is going to be zero is going to be one half times four times 10 squared, plus a final velocity of 40. And this is going to be for until t equals 15. So 15 minus 10 is five, two plus five times 40. So Delta s for car A is one half times force. That's too times 100 plus five times 40 is 20 times 10 is 200. So this is two times 200 which is just 400 m. At T equals 15 and for Blue Delta s and one half five times the time it's accelerating for which is five seconds squared. Plus, that's final velocity of 25 m per second time to the time that it's going. That velocity which is 10 seconds so that will give us five cubed is 125 So 125 over to plus 250 which we can simplify this as 1 25/2 is 50 plus 25 or two which is 12.5 plus 2 50. So this is going to be 300 plus 12.5, which is just 300 in 12.5. So that's our answer for adults A s and finding the difference between them is just 400 minus 312.5, which is going to be 112.5 was gone. 10 there nine. That's not a nine Sophie and eight. It's more like being paid, but we need a 0.5, so the distance between the two is 87.5 meters, and that's the solution.