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Q1,. In multi storey building, two elevators A and B in the adjoining shafts start moving in opposite directions with constant accelerations 0.2 m/s? and 0.4 m/s? r...

Question

Q1,. In multi storey building, two elevators A and B in the adjoining shafts start moving in opposite directions with constant accelerations 0.2 m/s? and 0.4 m/s? respectively. A moves down while B moves up. If they were 100 m apart at t = 0,after what time they will be opposite to each other? How far each one has travelled? Suppose both elevators are moving with same acceleration (say 0.4 m/s?) in the above case, what will be the effect on distance travelled and time? Critically think and discu

Q1,. In multi storey building, two elevators A and B in the adjoining shafts start moving in opposite directions with constant accelerations 0.2 m/s? and 0.4 m/s? respectively. A moves down while B moves up. If they were 100 m apart at t = 0,after what time they will be opposite to each other? How far each one has travelled? Suppose both elevators are moving with same acceleration (say 0.4 m/s?) in the above case, what will be the effect on distance travelled and time? Critically think and discuss. 8 Marks)



Answers

(II) 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 $1.2 \mathrm{m} / \mathrm{s}^{2},$ and $(2)$ the typical maximum speed
attainable is about 9.0 $\mathrm{m} / \mathrm{s}$ . You board an elevator on a skyscraper's ground floor and are transported 180 $\mathrm{m}$ above
the ground level in three steps: acceleration of magnitude
1.2 $\mathrm{m} / \mathrm{s}^{2}$ from rest to 9.0 $\mathrm{m} / \mathrm{s}$ , followed by constant upward
velocity of 9.0 $\mathrm{m} / \mathrm{s}$ , then deceleration of magnitude 1.2 $\mathrm{m} / \mathrm{s}^{2}$
from 9.0 $\mathrm{m} / \mathrm{s}$ to rest. (a) Determine the elapsed time for each of these 3 stages. $(b)$ Determine the change in the magnitude
of the normal force, expressed as a $\%$ of your normal weight
during each stage, (c) What fraction of the total transport
time does the normal force not equal the person's weight?

In this part, we need to calculate the distance trip we will use below formula As is equals two U. T. Place up. It is where? Yeah. Let us substitute the given venue. So you is given us to Metal persons. It's two multiplied by time. It's given us 0.2 2nd. We have calculated in previous problems plus of excellence and we have found us Excellence in is given a 0.2 metaphors and square multiplied 20 m30.2. Yeah. So after resolving we will get, the distance traveled is equal to 0.404 metre. Now, let us say for the options. Obviously see Dobson is correct.

But here, in this question we need to find out time taken by the stone to return the floor. So people right? The relative acceleration first so a relative Will be equals two 9.8 G. Does the even acceleration at a 0.2 it will be equal strong 10 m/s squared. No, you can find out the name of late t. V. Equals to do you scientists who your relative scientist to where did where? Really It will be equals two woman prepared way. Your relative has given us to plant a tree to has given us 30°.. But it was a relative term. So from here we will get the quiet time of life will be close to who would ever invented then, But it will be equals to 0.2 seconds. The knowledge is sick for the options. Obviously obscenity is good.

So here in this case we have been asked to find out it's time of light inside the lift. So as we know, time of flight, His return is T. Z. equals two to you, divided by G. Interactive. Oh here. In this case we have been given that inside of which is moving vertically upward with the uniform excellence and a body is projected vertically upward with speed you with respect to live. So we fortunate to calculate the g effective or the net effective acceleration. So we can write it as to you divided by a factory net active acceleration. So in this case the relative acceleration or the net acceleration Will be equals two. A plus G. A. Policy since since both are going upward So we can write it as the z equals two. Do you divided by a place? This is the required time. A plate. Let us not check the options. We see. That option is correct.

For birthday of this problem, we need to use Newton's second law to find the answer. But first let's draw a free body diagram to visualize the scenario here we have the elevator and we have the tension force going up and we have gravity going down. Now let's think about this scenario where we would have the maximum amount of force. Well, if you want the maximum then tension force has to be greater than the first of grabbing. And we know that it can accelerate upwards. So we can say that the maximum force of tension will occur when acceleration is positive. Mhm. If we define that positive, it's um the negative is down. Yeah. So what we can write is the mass of the elevator and the people times the acceleration is equal to the force of tension minus the force of gravity, which is mass times right? And now we get that mass times acceleration closed robbery users potential force over here all the day was fact rob the mass since we have it in both terms. Now, we can simply substitute the information, we know that this is how much we have for the elevator and the people inside of it. We know that we are actually reading upwards with 1.2 m per second squared and then we have grabbing and when we put this in a calculator we got this much force great. Now we have the maximum forest. This is imagine now that's think about the meeting. You might be tempted to say well when the elevator is not moving then the force of tension must match the force of gravity so that is the minimum. But actually the excellent. If the elevator is accelerating downwards then the force of tension will be less than the force of gravity. So now we have that the minimum happens when the force of tension, the minimum in the first of tension happens when the acceleration is negative. So we'll have a negative mass times acceleration is equal to the force of tension minus the force of gravity. So we get back, the force of tension is equal to the mass times gravity minus the acceleration. And again we have all the information we need. We simply substitute. Yeah. And we get that. The minimum force is roughly 6000 newtons less, So it will be 20604. Yeah. And this is an army. Yeah. Great. Now let's move on by me here. We can use our cinematics equations and notice that we are being asked for the minimum time. So that means that we want to make the trip as fast as possible. Now we also have to remember that the problem told us that we cannot go up faster than 18 m per second. So that means that we'll have to split the motion into first. We'll have acceleration going up of 1.2 millions per second squared, and then we'll have constant velocity of 18 m/s. This first part happens in a time T. one and the second one happens any time T to So let's set up an equation to represent the entire motion. The total displacement, the height will be equal to first, the displacement in the first portion of the motion. So that is going to be one half times the acceleration Times. Time one Squared plus the second portion. So that is going to be V times T. Sure. Okay now we have two unknown. So we cannot simply solve for this equation. But what we can do is find another way to solve for T. two. We know that final velocity is equal to the acceleration times the time. So we can calculate T one by knowing that the final velocity has to be 18 m/s. So T one will be the final velocity Over our maximum acceleration of 1.2 m/s squared Which gives us a time of 15 seconds. And now we have all we need to solve for T. two. Yeah. So where we arranged the equation and then we have to divide the entire equation. Bye baby. So we got T. Two. It's equal to this. And now we simply substitute the numbers. Mhm. Which gives us the time of roughly 28 point one. Now that's T. two. We need to find a total time which will which will be T one bus T two. Yeah. Yeah. Here we have time one Yeah. And one and 2. And we get Mhm 43.1 seconds. That is our final answer. Well, let's move on to Part C. In party we have a passenger. So let's draw everybody diagram of that passenger now that passenger is going through uh force of gravity downwards and the force of the ground. So, normal force going up now according to Newton's second, but we can say that mass of the passenger times the acceleration will be equal to the normal force minus the force of gravity. And we want to find the maximum and minimum. Well, the maximum is when the forest of the ground not of course is greater than the force of gravity. So we have upwards acceleration. So let's write that down over here. Maximum normal force is when we have positive acceleration. So that will be the mass times the acceleration class. The force of gravity and that will be equal. So the normal force. So normal force will be 60 kg times one point to the maximum acceleration the elevator can have plus gravity and that gives a force of 660 mutants. Now let's talk about a minute. Yeah, the minimum has to be again when the normal force is less than gravity. So acceleration has to be downwards. A minimum right acceleration is native. So let's rewrite these equate the situation with that mm So that is the normal force is equal to the mass things property minus the acceleration. That will be 60 kg times 9.8 minus 1.2. Yeah, that will be yeah, 516 m. So yeah, we have 516 Newton's and that is our minimum ports. Now let's move on to part the 40. We have the exact same scenario as we did in part B, but we have different values because the mini the maximum velocity that we can have downwards is no longer 18 10 m per second. Mhm. So we can use the same. And yeah, that would get up here, we have a height. Yeah, We have the first portion that can accelerate at 1.2 m/s squared and then we have time one square plus the velocity of decent times. Time to Yeah, so let's follow what we did in part because that part helped us find These two times. So that's fine. T to us we get the ball we have final velocity is equal to the acceleration. Thanks Time one. Oh sorry here I underline the wrong time. Now we'll find time one. So we have gravity times. Time one. Therefore time one will give us velocity over Right, So that is those 10 m per second over 9.8 m per second squared, That is roughly 1.02 seconds. So now let's use this equation that we found up here to solve for time to So let's take it down with Yeah, we have heart time to which will be equal to the height. And the height is the same as about Because we're going to the observation that So that is 640 m minus one half times the acceleration of 1.2 m per second squared times the time we just found. Yeah, That is 1.2 squared all over. The constant velocity of 10 years per second. Mhm. That gives us a time. Well, probably 63.94 seconds to find a total time. You will at 1.02 seconds plus This 63 94 seconds, That will give us 64 seconds 0.96 And that is our final answer.


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