High-speed elevators function under two limitations: (1) the m...

Question

High-speed elevators function under two limitations: (1) the maximum magnitude of vertical acceleration that a typical human body can experience without discomfort is about 1.2 m/s², and (2) the typical maximum speed attainable is about 9.0 m/s. You board an elevator on a skyscraper's ground floor and are transported 180 m above the ground level in three steps: acceleration of magnitude 1.2m/s² from rest to 9.0 m/s, followed by constant upward velocity of 9.0 m/s, then deceleration of magnitude 1.2m/s² from 9.0 m/s to rest. Determine the elapsed time for each of these 3 stages.

Accepted Answer

Welcome back. Everyone in this problem. A service elevator in a large hotel is designed to transport equipment vertically with minimal discomfort to the operator while ensuring efficiency. The elevator's movement from the ground floor to a maintenance level. 120 m above involves three distinct phases. First, it accelerates from rest with a constant acceleration of 1 m per second squared until it reaches its cruising speed of 5 m per second. Next, it then travels at this constant speed for the majority of its vertical journey. And finally, it decelerates to a constant rate of 1 m per second squared until it comes to rest at the maintenance level. We want to determine the elapsed time for each of these three stages of the elevator's ascent. For our answer choices, A says T one equals four seconds. T two is 20 seconds and T three is four seconds. B says it's five seconds, 19 seconds and five seconds respectively. C five seconds, 18 seconds and five seconds respectively. And D 5.5 seconds, 17 seconds and 5.5 seconds respectively. Now to help us better understand this whole movement and the acceleration, let's try to represent it on an acceleration time graph just a quick sketch to visualize the three phases. So here set about acceleration and our time let me make the time axis a bit straighter. OK? Then we are saying that in phase one, our elevator accelerates from rest with a constant acceleration of 1 m per second squared until it reaches its cruising speed of 5 m per second. So we could represent it looking something like this. OK? Because it's accelerating at that rate. Next, it says it travels and let me just show that this is phase one here. Next, it says that it travels at a constant speed for the majority of its vertical journey. OK. So that would be represented as a straight line on our graph because the acceleration is zero and that's phase two. OK? And then finally, it says that it decelerates at the same constant rate of 1 m per second squared until it comes to rest at the maintenance level. OK? So again, we can represent that acceleration on our diagram as stage three. And now what we are trying to do is trying to figure out the time for each of these stages. In other words, T one, T two and T three. So to figure that out, let's try and use our kinematic equations to help us to look at what's happening at each phase. So let's start with phase one. OK? Now if we're talking about phase one right phase one is our acceleration phase. And if we're going to figure out the time, OK. The time t one, what we don't know, let's think about all the information we do know. So first of all, we know that it's accelerating at 1 m per second squared, we know that it does that until it reaches a, a speed of 5 m per second squared. So our final speed is 5 m per second, sorry, not 5 m per second squared, but we're talking about velocity. So it's just 5 m per second and it initially started from rest. So our initial speed equals 0 m per second. So let's think, do we know anything that relates time acceleration and the final and initial speeds? Well, we know that let me put that in red. OK. We know that final speed V equals U plus A. So we can rearrange this formula to solve for T because no, this means A T equals V minus U and thus T which is T one, the time in the first phase is going to be equal to V minus U divided by A. In this case, that's going to be 5 m per second minus 0 m per second, divided by 1 m per second squared. And that's gonna be 5 m per second divided by 1 m per second squared, which would leave us with five seconds. So the time elapsed during the first phase of acceleration is five seconds. Next, let's think about the time elapsed in the second phase. OK. Let me represent that here. Just show this line to separate them. And let's think about the time in the second phase T two. So let's go to phase two and during this phase is a phase of constant speed. Now, what do we know about this phase? Well, we know we're trying to find the elapsed time T two. And based on the information we are given it's a constant speed. So the acceleration is zero, right? And the final speed equals the initial speed. Ok, which is 5 m per second. So that's just the speed, 5 m per second. Now, how can we figure out the time? Well, just by looking here, it's not easily apparent what the time is going to be. So we'll have to try thinking in another direction. What else do we know about speed and time? Well, recall, OK, recall that speed equals distance divided by time. So time equals distance divided by speed. In this case, D two divided by V. Now, we have that speed, but what we don't have is our distance. Ok, that's what we don't know, we don't know the distance traveled in the second phase. But is there a way to figure that out? Well, what do we already know? We know that the total distance the total distance traveled is going to be a combination of the distances traveled in the three phases, D one, D two and D three. So we know our total distance is 120. And to find our second distance, we can subtract the distance in the 1st and 3rd phase from our total distance. No, again, we are on another, uh we've met up on another problem here because we don't know what the distance is in the first or in the third phase. But there's a way we can figure that out. There's a, there's a, an equation that relates acceleration to initial speed and to time, we already know the time for the first phase. So we can solve for our first phase or rather solve for the distance in our first phase. So we can use the kinematic formula that says S equals UT plus half A T squared. In other words, the distance is going to be equal to the initial speed multiplied by the time plus half of the acceleration multiplied by the square of time. So in this case, S which is D one is going to be equal to zero, multiplied by the time five seconds plus half of our acceleration, 1 m per second squared multiplied by our time squared. So that's five seconds squared. OK. No zero, multiplied by 50 and a half of one multiplied by five squared is going to be equal to 12.5 m. So the distance traveled in the first phase is 12.5 m. Now, how about the distance in the third phase. Well, there's something interesting about the third phase that our, our diagram will help us with as well. Because notice that in our first phase and our third phase, both of them have an absolute value for the acceleration to be 1 m per second squared. So since the 1st and 3rd phases are symmetrical, the total distance covered during those phases are the same. In other words, the distance in the first phase equals the distance in the third phase. So that means D three is also equal to 12.5 m. OK? These three is also equal to 12.5 m. I think we can put that in a square here. So since we have those distances, OK, then that means I think I can make some room since we have those distances, that means let's make some space over here. That means D two, OK? Is going to be equal to our total distance 120 minus 12.5 plus 12.5 which is 25. So that's 120 m minus 25 m which equals 95 m. Know that we have the distance covered during the second phase we can solve for the time because that means the time taken is going to be equal to that distance 95 m divided by the speed 5 m per second. And that is going to be equal to 90 in seconds. So the time in the second phase is equal to 19 seconds. Finally, let's think about the time in the third phase. So let's make again a little separation here. And in the third phase, that is our phase for deceleration. Not remember what we said earlier, the time to decelerate should be the same time as the time to accelerate. Since the acceleration and deceleration rates are the same. And the process is symmetrical, which is what helped us to see that their distances were the same. So that means T one or rather T three time in phase three is going to be equal to the time in phase one, which was five seconds. So that tells us then that the time in the first phase is five seconds, the second phase is 19 seconds and the third phase is five seconds. If we go back to our answer choices that shows us B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Acceleration

Constant Velocity

Deceleration

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