You are 9.0 m from the door of your bus, behind the bus, when it ...

Question

You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s². You instantly start running toward the still-open door at 4.5 m/s.b. What is the maximum time you can wait before starting to run and still catch the bus?

Accepted Answer

Hey everyone. Welcome back in this problem. A train leaves a station terminal with a constant acceleration of 2m/s squared. A late traveler five m behind the entry of the train runs at a constant speed of five m per second. Trying to access the train were asked, how long can the traveler wait before we're beginning to pursue and still make it to. The train were given four answer choices. We have a 0.25 seconds, be one second, C 2.5 seconds and D five seconds. Now let's go ahead and write out the information we've been given. So we have information about the train and we have information about the traveler. So for the train it's going to leave the station. So its initial speed is going to be zero, it's going to be stopped at the station and then it's going to continue moving. So the initial speed is zero m/s. We aren't given information about a final speed. We're told that it has a constant acceleration of two m per second squared. And when you hear constant acceleration, you should be thinking uniform acceleration, which means are you A M uniformly accelerated motion equations or cinematic equations as they're sometimes called A, that means we can use those equations. Alright, the distance the train travels, we aren't told any information about the distance of train travels, but I'm going to come back to this. Okay, because there is a relationship between how far the traveler is going and how far the train is going because the traveler is starting behind the entry of the train. And then we have our time t okay. When we talk about the traveler, The speed of the traveler is 5m/s and they're going at a constant speed, which means that we only have five variables to worry about. Hey, their speed is five m/s. We're gonna call their distance D and the time T, well, this is gonna be the same. I'm just gonna write T equals T, which is a little bit redundant, but I just want to make the point that these are gonna be the same time. We want the traveler in the train to meet at the same time point. And in order for them to meet the train has to have traveled a distance D minus five m, 5 m less than the traveler. Since the traveler started five m behind the entry of the train, they have to travel five m further. And you could do this either way we've chosen to put the D minus five for the trains displacement. You could also put D as the distance the train travels and then D plus five for the traveler, whichever way you want to do it, it's gonna work out just the same. Alright, so we have four pieces of information about the train. We don't have values for all of it, but we have four pieces of information. So let's try to write a you Am equation with those four things that we have information about. So we're gonna ignore V F. We're gonna choose the equation without VF and we get that the displacement D T, it's going to be equal to be not multiplied by T plus one half multiplied by a multiplied by G squared. And we can substitute in the information we know. So we have D -5 m is equal to V naught zero. So this entire first term goes to zero one half multiplied by two m/s squared multiplied by T squared. And if we want to write this just as D is equal to, we have D is equal to, we move the five m to the right hand side, we get one m per second squared, T squared plus five m. All right, let me put this in red. Okay. This is the equation that's going to govern the train. We're gonna do the same for the traveler. Okay. Now the traveler is moving at a constant speed. And so the equation that governs that motion is the simpler equation where speed V is equal to distance over time. So if we want to write this as a function in A D, in terms of the other variables, just like we did, we can write that the distance D is equal to the speed V multiplied by time T. And in this case, we get D is equal to five m per second multiplied by the time T, we're gonna put a blue box around this one. So the blue is going to represent the equation for the traveler. Okay. So you'll notice that this equation for the traveler is if they start moving right away, we're trying to figure out how long they can wait. Let's go ahead and just draw a diagram. Okay. We're gonna draw a little graph, a sketch of these two functions that we have and try to figure out what's going on. So we have our X axis which is going to be representing time. We have our Y axis which is going to represent the distance D. Now our function D is equal to T squared plus five m. This is a quadratic function And it has a y intercept of five we say to. Alright. So this is going to look something like the following. That is that function D is equal to one m per second squared T squared plus five m. We can do the same for our blue equation. Now this is going to start at zero and it's gonna be a line and it's gonna cross. Oh, that's not quite a line. All right, you can let me try once more to get this looking nice. Alright, so you can imagine that's a nice straight line. And what we have is we have an intersecting at two points. If we start right away, if the traveler starts right away, there's gonna be two time points when they can catch that train. Now we want to figure out how long they can wait. What we can imagine is that if the traveler waits, okay. It's effectively gonna shift this line, this blue line to the right horizontally, okay? Because if they wait then the initial time they start at is a little bit later. Okay. So you can imagine shifting the line stay out here and if they wait this long, if we shook the curve too far to the right, then the line in the Parabola are not going to intersect and the traveler will not be able to make that train. Alright. So when we think about the longest possible time, they can wait. Well that's gonna be if we shift the line oh, until There is only one intersection. Okay. So there's only one intersection point. If we shifted any further to the right, they're gonna miss the train. So that's the longest possible time we can wait and that's gonna be this time here. We'll call it T W T wait. Alright. So are shifted equation. If we want to write an equation for this new line we've drawn, That's gonna be D is equal to 5m/s, multiplied by the time TK has the same slope as the original equation. But we have this extra initial point that's not. And right now, we don't know what Exxon is. We're going to try to find it. Now, in order for the traveler to catch the train, we want the distance to be the same. Okay. We have D which we've said is the same for both. So we want those to be equal, we can set them equal to each other and we get one m per second squared times T squared. Last five m is equal to five m per second, multiplied by T plus X not. And you'll recognize this is an equation of a second degree polynomial. We have a T squared term, a T term and then constant terms. So I'm gonna drop this one m per second squared. The coefficient is just one. I've left it there to indicate the units. But for right now, I'm just gonna drop that the units there. So we can see what's going on with this function. If we really reach and put everything to one side, we get T squared -5 m/s, multiplied by T plus five m minus X not is equal to zero. Alright, so we have this quadratic function that governs the intersection of the travelers distance and the trains distance, we want there to be only one intersection. And recall that if we want one in our section for a quadratic equation, This means that the discrimination should equal zero. Alright, now recall what the discriminate is. Okay. So the discriminative we'll call it capital T is equal to B squared -4 multiplied by a multiplied by C. Now, in our case B is going to be the term the coefficient of the T term. So it's going to be negative five A is going to be the coefficient on the T square term which is one And C will be the constant term, which is five m -X not. So if we send this into our discriminate, We have that negative five and I'm gonna drop the units just for a second. We have negative five squared minus four, multiplied by one multiplied By 5 -1 not equals zero. Okay, we can solve for X dot Now we have zero is equal to negative four times five gives us negative negative four times negative X not gives us positive four, X, not, we get that four X not is equal to negative five. And here we had meters per second, all squared R unit. Here it's meters squared per second squared. And on the value of four, we had meters per second squared. So X not will work out to be in meters and we get a value of negative 1.25 m. Alright. So in order for there to be only one intersection point, which means we waited as long as possible to catch that train, we need X not to be equal to one, negative 1.25 m. If X is equal to negative 1.25 m, then the equation governing the traveler. Okay. And I'm going to read it in blue because we were writing it in blue above is going to be d equal to five m per second, multiplied by T minus 1.25 m. That value of X not we just found. All right, let's look at our graph here. Remember we're trying to find this waiting time. The waiting time is going to be the X intercept the time when the distance is zero. Mhm. So in order to find the waiting time, we set the distance equal to zero, that means the traveler hasn't moved yet. So they're still waiting. We got five m per second times. We're gonna call it T w that waiting time -1.2, 5 m. And we get a waiting time tw Of 0.2, five seconds. Mhm. All right. That's it. We made it to the end. So that was a long problem. But we just worked through it piece by piece using those Kinnah Matic equations and using what we know about quadratic equations as well. And we found that the traveler can wait 0.25 seconds before starting to move and they will still make that train. So when we look at our answer choices, this corresponds with answer choice. A 0.25 seconds. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Physics?

You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s². You instantly start running toward the still-open door at 4.5 m/s.b. What is the maximum time you can wait before starting to run and still catch the bus?

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