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A race driver has made a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is $6.0 \mathrm{m} / \mathrm{s}^{2} ;$ after 4.0 $\mathrm{s}$ he enters the main speedway. At the same instant, another car on the speedway and traveling at a constantvelocity of 70.0 $\mathrm{m} / \mathrm{s}$ overtakes and passes the entering car. The enteringcar maintains its acceleration. How much time is required for the entering car to catch the other car?

$t=15.33 \mathrm{~s}$

Physics 101 Mechanics

Chapter 2

Kinematics in One Dimension

Motion Along a Straight Line

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in this problem. We're looking at a race car that begins its path in a pit stop, and as soon as it leaves, it is racing to catch up to another car that is not accelerating. So let's kind of set this problem up. Um, over here on the left, I'm going to have the initial be kind of rev up period rev up. And then once it leaves the pit stop, we have the first car, which is the one we're tracking, as well as the car with no acceleration, that that car is trying to catch up. And this problem is essentially asking us. How long will it take our first car to catch that second car once they both get out onto the track. So let's start with what we know. Um, first of all, because our initial car starts at a pit stop, it starts at rest, so its initial velocity is going to be zero meters per second. Now. We also are told that the acceleration is six meters per second squared and that the car drives with this acceleration for four seconds. Now, once it leaves that pitstop area, it continues to accelerate at six meters per second squared and the car it's racing with has no acceleration. So it's trying to catch this car that is moving at a constant velocity, and we know that that constant velocity for that car is going to be 70 meters per second. So a piece here that connects all these problems is that the rev up period leads directly into what we have on the right hand side here with the first car trying to catch that second car. So the final velocity from the pit stop, that final speed it gets up to is going to be the initial velocity for the second part. So before we figure out how long it will take the cars to catch, we need to figure out what that final velocity for the rev. Up period is going to be now. The other pieces we know that I want to spell out before we get to solving this unplugging in numbers is that when we're looking for how long it takes the first car to catch that second car, the time is going to be the same for both of these things because it will take Car one. The same amount of time. It'll take our two for them to intersect. Then the distance they travel is also going to be the same thing because we're looking for that moment in time where they're both at the same position. So when we start solving things, we know those values are going to come out to be the same. Okay, so with all of that set up and down, let's jump in to how we actually solve this. Our first step is working with that rev up period, and in that case, we're trying to find the final velocity now for looking at what else we know from that problem. We do not know the distance, and we also don't need to know the distance of that rev up period. So we're going to use the equation that does not have an X, which is this first equation, which I think is everybody's favorite equation because it's easily the most straightforward. And so when we get to plugging that in, call this step one, we have V equals V not plus 80. When I plug in my numbers from that rev up period, we found that the final velocity is equal to zero plus six multiplied by four. So it's going to mean our final velocity is 24 meters per second. So move this out of the way. There's step one done. We know that final velocity is going to be 24 meters per second so that initial velocity is also 24 meters per second. Now, Car one and cartu, we kind of have to do together because neither side of these do we know anything. We don't know three numbers for certain about either one of them. But the fact that they're teaser the same and their exes air the same means we'll be able to actually solve this and work through it. So our goal here is to figure out how long it takes them to overlap. And we know that they're ex values air the same. So this means that we don't need to use their final velocity when we're working through these problems. That number does not matter for these. So for that first cars velocity are the first cars Cana Matics Here. I'm going to use the equation X equals V, not tea plus 1/2 a T squared. Yet I'm using that one because it does not contain the final velocity. And the final velocity is the peace. We don't know anything about here. Now we don't know what X is. We know the initial velocity is 24 but we don't know the time it takes to catch the other car. The acceleration is 6 to 1 to 1/2 the acceleration. We get three t squared. Okay, then the second car, We're going to use that same equation for but since the acceleration is zero, it actually becomes a pretty short problem. You know, change the color to be that one as well. So we end up with X is equal to 70 because that's how fast that other car is traveling, multiplied by the time I don't need to worry about the axe of a or the T of a because x ntr the same for both cars, which means I'm gonna be using the substitution principle here to solve for me. One trick, though that is pretty nice for this problem is if we divide the tea out on this side, we get X divided by T is equal to 70. If I do the same thing for my first problem and divide a t out. We get X divided by t equals 24 plus 32 or three t then using substitution. So I'm gonna plug my 70 in here we end up with 70 equals 24 plus three t so 46 equals three times the time. So my time, The period where these two cars will overlap where the first car catches and then passes. The second car is 15.33 seconds.

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