At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 2.80 m/s2. At the same instant a truck, traveling with a constant speed of 20.0 m/s, overtakes and passes the car. (a) How far beyond its starting point does the car overtake the truck?

Video transcript

Welcome back everybody, We are taking a look at two competing delivery trucks here, we're gonna start out with the delivery truck for the swift company and we are told that it starts from rest and then just starts driving. Now at the instance that it does start driving, there is another delivery truck for the delivery company, Thunder and it starts driving the same direction. However, at the instance that Swift starts driving the thunder truck overtakes the swift truck. Now we are told a couple of things about this entire system here, we are told that the initial velocity of the swift truck is zero, we are told that its acceleration is 3.2 m per second squared. We're told that the constant velocity of the thunder truck is 22 m per second and this point right here where it kind of just starts out where the thunder truck overtakes the swift truck, we're just going to label that as the initial position of zero. Now, eventually there is going to be a point in this system where the swift truck will then overtake the thunder truck, so it will be in the lead this time where the swift truck, I'm gonna label this point X. S. And for the thunder truck, I'm gonna label it X. T. But you can see that both of these Xs are the same and we are tasked with find finding what this X. Is. Now, what is the easiest way to achieve it. Well, let's look at this equation here, I'm gonna look at it, I'm just gonna call equation one, we know that the position of a given object is just the velocity over time. And conveniently we are already given the constant velocity of the thunder delivery truck. So I'm just gonna say that we should find X. Of T because S is equal to X. Of T, meaning that this is gonna be equal to VT times time. But what is time here? We're going to have to use another equation which I will call equation too, that x minus X not is equal to V naught times T plus one half a t squared. Now I'm gonna add X not to both sides of this equation. Council's out on the left hand side and we are left with that X. Is equal to all of this. Now this X right here can both be X. S and X. T. So if we find Xs and XT with this equation, we'll be able to solve for time and then plug that into the first formula. So let's go ahead and do that here. X. S. Is going to be equal to X zero. Our initial position of zero plus our initial velocity for the swift truck which is zero times time plus one half times 3. T squared. This just gives us 1.60 squared great. Now what about X. Of T X. Of T. Is going to be our initial position of zero plus our velocity. Now it does say initial velocity here. But if we have a constant velocity that means two things. It means one, that the acceleration of our our thunder truck is going to be zero and two. It means that the final velocity equals the initial velocity. So we're allowed to plug it into here. So I'm gonna plug in 22 here times time plus one half times the acceleration which we just said with zero times T squared. This gives us 22 T. So now what I can do is I can set these two things equal to each other and just label it X. Right, so we have X. Of T. Is equal to our X. Of S. Now this gives us the following equation. We have 22 T. Is equal to 1.60 squared. I'm actually gonna subtract 22 T from both sides which just makes this side zero. So now we have this quadratic equation is equal to zero. So in order to find what the possible values of T. Could be, we're just gonna use the quadratic formula. So we have negative B. 22 plus or minus the square root of B squared minus four times a which is 1.6 times C. Which is just zero in this case. So this entire term is going to go to zero all over two times A which is 1.6. This really just simplifies down to 22 plus or minus 22 all divided by 3.2, Which gives us two possible answers for time zero or 13.75 seconds. Let's think about this logically here. We know that some time passes between this point and this point, meaning our time can't be zero, meaning the only other choice it could be is 13. seconds. So now what we can do is we can take this time and plug this back into our first equation here to find our ex. Or our X. Of T. Here. Remember X. Of T. Is equal to X. Of S. Is equal to our desired X. Right? So let's go ahead and plug it in. Remember X. Of T is equal to V. T. Times T. Which this is going to give us 22 times 13.75. Giving us 302. m corresponding to our answer choice of C. Thank you all so much for watching. Hope this video helped. We will see you all in the next one.