At the instant the traffic light turns green, a car that has been...

Question

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. (b) How fast is the car traveling when it overtakes the truck?

Accepted Answer

Welcome back everybody. We are taking a look at two different competing delivery vans that are driving past one another. We start off with the Swift Company van that is driving down the street. Now, as soon as it starts driving a competing van for the Thunder Company passes it right as soon as it starts driving here. So I'm gonna label this little van with a T and it's driving in the same direction. Now we're told a couple of different things about this entire system here, we are told that the swift van starts off from rest and we are told that its acceleration is 3.2 m per second squared. We're told that the Thunder van as it's already moving, had a constant velocity of 22 m per second. We're also told that, let's just say at the time of thunder overtaking swift at the very beginning will just mark that position as position zero. Now, with all of these facts about this system, we are told that at some point the swift van will then overtake the Thunder van. Now, I'm gonna label this point X, S for the swift van and X of T for the Thunder van, but we can just see that those values are equal. Now, we are tasked with finding when the swift van over takes the thunder van, what is its velocity for this? We're gonna have to use a couple of kinetic equations here. Now, a direct equation for this velocity right here will be that the final velocity of our swift van at this point is going to equal the initial velocity plus its acceleration times time. Now we have both of these terms but we do not have time so we have to figure that out. And so I'm gonna label this equation equation one and then let's use a second equation to figure out our time. We have that the uh final position or these positions right here minus the initial position equal to the initial velocity of either van plus one half times the acceleration times time. I'm actually gonna add this initial position to both sides. And what we can do now is we can use this equation to solve for both X. S and X. T. And since they are equal to one another, that will allow us to solve for time. So first and foremost let's solve for X. S and X. Of T X. S. Using this equation right here is going to be equal to Well, our initial position is just zero plus our initial velocity is zero times time plus one half hour acceleration of 3.2 times times squared these two terms go away. So we're just left with one half times 3.2, which is just 1.6 times T squared. Great. Now, what about our X. Of T? What is this at this position? Well, our initial position is zero plus our constant velocity of 22 times time. Now notice how I'm using the constant velocity here, that means both the final and initial velocity are the same. So we can actually use that here plus one half times are acceleration. But if the velocity is the same, its acceleration is zero. So this is just going to be zero times T squared these two terms will disappear and then we are just left with 22 T. So now let's set X of T equal to our X of S. This gives us, Let's see, 22 T is equal to 1. T square. I'm gonna subtract 22 t from both sides, canceling this out and therefore making this equal to zero. Now, if you look at this equation here, we now have a quadratic equation so we can use the quadratic formula. So we have negative B which is gonna be 22 plus or minus the square root of B squared minus four times a 1.6 times C which is zero. So this entire term is just going to disappear all over two times 1.6. What this really simplifies down to is 22 plus or minus 22. All over Earth. 3.2. Now, this gives us two different answers here. We get a time of zero and a time of 13. seconds. But we know some time passes. So it's impossible for this time to be zero. Therefore the time it takes for this uh for these vans to reach this point is 13.75 seconds. So now what we can do is we can take this time and plug this into our first equation to find the velocity of our swift van at that point. So we have that the velocity is equal to our initial velocity plus our acceleration times time. Remember our initial velocity for the swift van is zero plus our acceleration of 3.2 times the time we just found of 13.75 seconds, this gives us an answer of 44 m per second, which corresponds to answer choice. D Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

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