A car starts from rest at a stop sign. It accelerates at 4.0 m/s²...

Question

A car starts from rest at a stop sign. It accelerates at 4.0 m/s² for 6.0 s, coasts for 2.0 s, and then slows down at a rate of 3.0 m/s² for the next stop sign. How far apart are the stop signs?

Accepted Answer

Hey, everyone in this problem during a workout, a sprinter initially at rest at point a accelerates for five seconds at a constant rate of 2m/s squared. Afterward, the sprinter maintains a constant speed for 15 seconds before he decides to decelerate at a constant rate of 4m/s. Squared. The sprinter stops at point B and were asked to calculate the distance from point A to B. The answer choices were given are a 175 m. B 187.5 m sea 200 m and D 212.5 m. Now, let's think about this problem. This is a motion problem. We expect to be using arc idiomatic equations or are you A M equations? Well, what do we have? We have this initial phase where this French is accelerating, then they maintain a constant speed and then they decelerate. So we have three stages of motion here that we have to consider. So the first stage We're gonna call it stage one. This is when the sprinter is accelerating, we're told that they are at rest initially. And so the initial speed or velocity is going to be zero m per second stage one, we don't know their final speed. We know that they accelerate at a constant rate of two m per second squared And that this acceleration lasts for five seconds and we aren't told anything about the amount of distance this front er covers here. So that's our first stage. Then we have our second stage And in the second stage, the sprinter maintains a constant speed for 15 seconds. Okay. So when we have a constant speed, we only have three variables to worry about velocity, distance and time and we know that they maintain a constant speed for 15 seconds. We aren't told any information about the distance they travel. We aren't told any information about their speed. However, we can relate the speed and stage two to the speed in stage one The final speed in stage one after the sprinter has accelerated is going to be that same speed we have in stage two that the sprinter maintains constant. Okay. So we have that V2 is actually going to be equal to VF- one and these are all velocities. So that's stage two. And then we move to stage three where the sprinter is decelerating our initial speed and stage three. Well, that's going to be the same speed that we stopped stage two in. Well, we had a constant speed in stage two. So this is equal to V two and we know that V two is equal to V F one. And so these three speeds or three velocities are all going to be related. Our final speed in stage three is going to be zero m per second because we're stopping decelerating to a stop the acceleration we're told is four m per second squared. And it's a deceleration. So it's going to be negative for meters per second squared. We don't know any information about the time this takes and we don't know the distance D three. Now, what we want to find is a distance from A to B A is the initial point where we start, this printer starts running and B is the final point where this printer stops running. And so the distance that we're trying to find distance from A to B, it's going to be the sum of the three distances from our stages. So we have the one Plus de two plus D three. So what we need to find is D one, D two and D three in order to find the distance we're looking for. So how can we do that? Well, in each of stage one and stage three, we have three variables. Okay. And let's actually start with stage one because we have three that we have numeric values for in stage one, we have V, not one, A one and T that's gonna allow us to find that distance D one we want to find Using that we can also find this speed VF1, which we can plug into our other two equations. Okay. So let's start on stage one where we have the most information. So starting with finding the distance D1, we're going to choose an equation that doesn't have VF in it because we don't have information about VF. And that's going to be the following that the distance. In this case, do you want Is equal to v? Not one multiplied by the time T plus one half multiplied by the acceleration. A one multiplied by the time T one squared, we know that V not one is equal to zero. And so V not one multiplied by T is going to be zero. The entire term goes to zero. So we're left with one half multiplied by two m per second squared, multiplied by five seconds squared. This is going to give us a distance D one of 25 m. We're gonna put a blue box around this so that we don't lose track of it. We're gonna need to come back to it in order to calculate our total distance. Okay. Now, that's great. We found our distance D one but in order to find our distance D two and our distance D three, we need to know this value of V two and we need to know the value V not three. Remember that those are equal to the final speed from stage one. So let's use the information in stage one to find that speed. So now we're going to choose an equation that includes VF one, we have all the other variables. So any of the equations that have V F one in it are Good choices. We're gonna choose the one that's just the simplest to calculate. So we're gonna choose VF one is equal to V not one multiplied, oops, sorry plus a Multiplied by the time T one again, be not one is zero. So the first term goes away, we have the acceleration two m per second squared multiplied by the time five seconds. And this gives us a final speed for stage one of 10 m per second. Now, we can use that speed in our other two stages. So let's move on to stage two. In stage two, we have a constant speed and let me label this as stage one. So we don't get confused and we do have the subscript. So the subscript indicate what stage we're talking about, but let's just group it. So it's clear. So again, stage two has constant speeds when we're talking about an equation that governs this motion. What we have is that the velocity V two is equal to the distance D two or the displacement D two divided by the time T two, We know that the speed is 10 m per second because it's the same speed as the final speed of stage one. This is equal to the distance D2 divided by the time. And in stage two, we spend 15 seconds. So we get D two is equal to 10 m per second, multiplied by 15 seconds, Which gives a value D two Of 150 m. So we've got D one, we've got D two, we have one more value to find before we can find that total distance. And that is going to be D three, that distance that we travel during stage three, the deceleration phase. So this is stage two that we've just done constant speed phase and moving to stage three, let's go back up to the variable. So we can see what we need. We know V not three now because we calculated VF one, that's 10 m per second. We have a VF3, we have the acceleration, we're looking for the distance. And so we're gonna choose the equation that doesn't include the time teeth, okay. We don't have information about the time T and that's not what we're looking for. So we are going to choose that equation and that is going to be V F squared. And in this case via three squared is equal to V, not three squared Plus two, multiplied by the acceleration Multiplied by the Distance Day three. So our final speed is zero. So we have zero on the left hand side is equal to 10 m per second squared plus two multiplied by negative four m per second squared times the distance D three, We want to isolate D three. So we can move this term to the left hand side. We get eight meters per second squared Times D three is equal to meter squared per second squared. And dividing by our eight m per second squared, we get D three is equal to 12.5 m. Okay. So we have our D one R D two R D three values. All that's left to do is some them together. Okay. Let's go back up. Remember that we're looking for this total distance from A to B and that's going to be the sum of those three distances. So our total distance from A to B Is going to be equal to 25 m Plus 150 m Plus 12. m For a total distance of 187.5 m. And we've got that value we were looking for now. So 187.5 m. If we go back up to our answer choices, We see that this corresponds with answer choice B. We found that the distance from a to B is 187. m. Thanks everyone for watching. I hope this video helped see you in the next one.

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A car starts from rest at a stop sign. It accelerates at 4.0 m/s² for 6.0 s, coasts for 2.0 s, and then slows down at a rate of 3.0 m/s² for the next stop sign. How far apart are the stop signs?

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