A hotel elevator ascends 200 m with a maximum speed of 5.0 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s².

b. How long does it take to make the complete trip from bottom to top?

Question

A hotel elevator ascends 200 m with a maximum speed of 5.0 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s².

b. How long does it take to make the complete trip from bottom to top?

Accepted Answer

Hey everyone in this problem, we have the operation of a winch divided into three successive stages. We have the acceleration period where a one is equal to 0.5 m per second squared We have the constant speed period where V is equal to 2.5 m/s And then we have the deceleration period where a three is equal to negative 0.5 m/s squared. A worker uses a winch to pull vertically stacked crates from the ground level and were asked to determine the time required to pull a crate m. Alright, so in order to pull a crate, 45 m, we're going to have to go through all three stages. Okay, we're going to have to go through some acceleration period Until we get to our constant speed period and then we're also going to have to decelerate and so we're going to need to figure out the time that we spend in each of those periods in order to move our crate m. So starting with the acceleration period, Okay. And this is gonna be our first stage. So we're gonna use subscript one, just like the problem did when they gave us a one. Now, our initial speed here, we're moving these crates from rest and so the initial speed of the crate is going to be 0m/s. The final speed. Well, what's the final speed going to be? We know we have the acceleration period followed by the constant speed period and the speed and the constant speed period is 2.5 m per second. So this means that our acceleration period is going to last until we hit the speed, the f one of 2.5 m per second. The acceleration were given in the problem 0. m/s squared. We don't know the time that this takes and we don't know that distance from the ground or the delta Y. That this moves the great. And what we want to do is we're gonna need to figure out both of these things. Okay, the question is asking us the time required so we do need to figure out T1 that's gonna be part of that total time. But we also need to figure out how much this period moves a crate because we need to move the crate. 45 m. Okay, so we're gonna have 45 m and whatever distance the acceleration period moves. The crate is going to be subtracted from that total 45 m going into the next two periods. Okay, so we actually need to figure out T one and delta Y one. So let's start with delta y. How how far does this acceleration period? Move the crate Now we're going to choose. Are you A. M. Equation that does not have time included. Okay, we don't have information about the time and right now we're trying to find information about delta y. Okay, we have three knowns so we can use these equations and we're gonna have VF one squared is equal to V not one squared plus two A one delta Y one VF one squared. Well this is going to be 2.5 m per second squared. V not one squared. Well that's gonna be zero. So we're just gonna have to times a one which is 0. m per second squared times delta Y one, two times 0.5 is just going to be one. And so we get the delta Y one is equal to we have meters per second squared. And on the left we have meters squared per second squared. This is gonna leave us with 2.5. Oops. And sorry, we have 2.5 squared when we square 2.5 we get 6.25 and we're left with the unit of meter. Okay? And so our delta Y one is 6.25 m. The acceleration period moves the crate. 6.25 m. How long does it take? Well, let's choose the um equation that doesn't contain delta Y one. So we're gonna choose V. F one is equal to V. Not one plus a one T one. Okay. And we actually know Delta Y1 now, so if you wanted to use any of the other equations involving T you could um we're gonna use this one. This is simple. Um and we're just gonna use the values we were given in the problem instead of the one that we calculated. All right, so we have 2.5 m per second. Okay. V not one is zero on the right, we just end up with 0.5 m per second squared times the time T one which gives us a time T one. Okay, 2.5 m per second divided by 0.5 m per second squared is going to give us five seconds. Alright, so acceleration period one lasts five seconds and it moves the block a distance of 6.25 m and I'm just gonna go ahead and write these in a red box because we're gonna need them later. Alright, so the next period is a constant speed period. The problem with the constant speed period is we don't actually know how long we spend in the constant speed period. We don't know how far the constant speed period moves. We don't know how long we spend there. Um So there's a bit of an issue there. So let's go to the deceleration period instead. Okay, like the acceleration period, we have a clear target of the speed that we start at the speed that we stop at and we will be able to figure out just like we did for the acceleration period, how far we moved the crate in that period and how long it lasts. So our deceleration and like they did in the problem we're gonna use subscript threes for this stage. Now our initial speed, we know our initial speed because it's going to be coming off of that constant speed period and the speed there is 2.5 m/s. So the initial speed in the deceleration phase is going to be 2.5 m/s. We are going to decelerate Until the block is not moving. So the final speed is going to be zero m/s And we know that our acceleration is negative 0. m/s squared. And just like for the acceleration period we want to know how far the crate moved. Once, we know that. Then we're going to know how far it needs to move in the constant speed period. And we're also going to need to figure out the time because that's what the question's asking. So we're gonna have to use two equations and we're going to use the exact same two equations we use for the acceleration period. So are you am equation without T in it is going to be V. F. And in this case subscript threes is equal to be not Plus two A Delta Y. So our final speed is zero Our initial speed is 2.5 m/s, all squared two times the acceleration, negative 0.5 m per second squared times delta Y three. The distance of that crate moves, that's what we're looking for. Alright. So if we isolate for Delta Y3, we're going to have negative 6. m squared per second square. Okay that's from the 2.5 m per second. All squared. We're gonna have two divided by negative one meter per second squared. And this is going to give us a distance of 6.25 m. So that's the same distance. So moving the block From the floor up 6.25 m We can do that the same in our acceleration period as our deceleration period. Let's get to the time and I'll show you what I mean there by doing this kind of the same way. So our delta y here 6.25 m finding the time again, we're going to use that same equation as we did in the acceleration period. V. F and in this case subscript three is equal to v not subscript three Plus A three T 3. Our Final Speed zero. Our initial speed 2.5 The acceleration negative 0.5 m/s squared. And the time we're looking and this is going to give us 83 value. Okay, we move this to the other side. We have 2.5 m per second divided by negative 0.5 m per second squared. And sorry I've said this is negative. Once we move it to the other side it will become positive. So we have a positive divided by a positive. Okay and this is gonna give us five seconds just like in the acceleration period. So in order to accelerate the block from rest up to the speed of 2.5 m per second. When we have an acceleration of 0.5 m per second squared, it takes five seconds. Now in order to decelerate the block from that same speed back to zero with the negative acceleration, negative 00.5 it's the same, it takes the same amount of time to accelerate up to a point as it does to come back down from that same point. Alright, so we've got our acceleration period, done our deceleration period done okay, We have two of the three times and we have two of the three distances. So now we need to figure out how much time we are in the constant speed area. Okay? Or face, so we know we want to a delta Y and this is going to be equal to the delta Y in the first phase plus the delta Y. In the second phase plus the delta Y. In the third phase we know we want delta Y to be 45 m. Delta Y one was 6.25 m. Okay, that's the acceleration phase. The crate moves 6.25 m. We want to know how far we have to move it in the constant speed phase, which is delta Y two And in the deceleration phase, Delta Y three, it also moved 6.2, 5 m. And so if we solve for Delta Y2, we're gonna have 45 m -12.5 m, which gives us a Delta Y two of 32.5 m. And so in our constant speed phase we need to move 32.5 m. All right now, in our constant speed phase. Okay, the speed is constant, that's given in the name. So we have no acceleration and so we can treat this as just a typical velocity equals distance over time problem. Okay, the velocity in Phase two is equal to the distance In Phase two over the time to In the distance D two. This is equal to Delta Y two over T two. and so what this tells us is that the time T2 we're looking for Is going to be equal to Delta Y two over the speed V. This is equal to 32.5 m at distance. We need to travel divided by the constant speed of 2.5 m per second and we get a time T two of seconds. Alright, so now we need to answer this question. The question was asking, how much time does it take to pull this crate? 45m and we had to find three times the time for the acceleration phase. The deceleration phase and the constant Speed phase. Okay, all three of those make up the 45 m. So if we go back up, We had five seconds five seconds and 13 seconds and the total time is going to be the sum of the three T one plus T two plus T three, which is going to be five seconds plus 13 seconds plus five seconds, which gives us a total of 23 seconds. Alright, So if we look at our answer choices, We found that the time taken to move this crate, 45 m was going to be be 23 seconds. Thanks everyone for watching. I hope this video helped see you in the next one.

A hotel elevator ascends 200 m with a maximum speed of 5.0 m/s. Its acceleration and deceleration both have a magnitude of 1.0 m/s². b. How long does it take to make the complete trip from bottom to top?

Verified Solution

Key Concepts

Kinematics

Acceleration

Uniform Motion