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Draw position, velocity and acceleration Vs. time graphs for both stones in the same coordinate axes given belowL0 timetime1time...

Question

Draw position, velocity and acceleration Vs. time graphs for both stones in the same coordinate axes given belowL0 timetime1time

Draw position, velocity and acceleration Vs. time graphs for both stones in the same coordinate axes given below L 0 time time 1 time



Answers

Sketch the acceleration-versus-time graph from the following velocity-versus-time graph.

For this problem, you have to refer to the graph it's given. Um they give you a position versus time graph and ask you to sketch the velocity versus time graph. I'm going to break up. The get graph is given in tow. 12345 different segments. You want to find the slope free because that will give you the velocity free to those segments. It's five different. Constant velocity is day. Graph transitions through so coordinates for the first segment 00 and 0.44 It's in meters and seconds. So changing whatever change in X, the slope or the velocity for this first segment is going to be changing. Why four minus zero over 0.4 minus zero. So four over 40.4 When you divide that out. Four. Divided by 40.4 i Z Will the 10 so 10 m per second positive for the first segment. Meters per second. That goes out 2.4 seconds. So it's moving at that speed for that amount of time. For Segment B, it starts at 0.44 and it ends at 0.6 negative too. So V would be changing. Why, negative too minus four over 0.6 minus 0.4. You get negative six over a point to divide that out. Sure. Open. And you get 30. Negative. 30 meters per second. That interval is from 0.4 seconds out. 2.6 seconds. See to keep track of that, Aziz going along here for C start to 0.6. Negative too. And it ends at one negative six. So the velocity is gonna be changing. Why? Negative. Six minus minus two. Over one, minus 10.6. So that is going to be plus positive. You're gonna get negative. Four over. 0.4 was going to be negative. 10 m per second. It's doing that from 0.6 out to 0.6. Out to one segment D one negative six. It ends at 1.6. Negative. Four changing. Why? For the sorry for the velocity you're going to do negative four minus minus six. Over negative. That should be negative. 1.6. Sorry. That's positive. Let me erase that real quick if I can. This negative does not belong. Yeah, so it doesn't belong here either. So 1.6 minus one, you get plus positive two divided by, uh 0.6. Do daughter by 0.6 is 3.33 meters per second. That is from, um, one out toe, 1.6. I should be a princess. It that's the time frame. And finally e 1.62 negative four 1.6. Negative four. Sorry to To To so v is equal to change them. Why? Um, this would be to minus minus four over two minus 1.6. You're gonna plus positive again. So this is six over 0.4 six. Divided by 60.4 is 15. Positive 15. That's happening from 1.6 seconds out to two seconds. So my graph, I've got a plot it out so that I could fit two seconds on their up to two seconds. Um, that means about halfway through is going to be one. Let's see if we could do it this way. So if I make this point to and go from there 4.4 point 6.8 one, 1.21 point 41.61 point eight should be too aan den vertically on velocity. I need to start. Let's see, I go down the negative 30 and up the positive. 15. So let's start up here somewhere. I'll draw in a line here. Has my access. Yeah, so let's go up. Let's see. 12 Let's see for five, 89 So I need to go up the positive 15. So let's just do it this way to keep it simple. 5, 10, 15, 2025. So, five for every space. So this would be negative. Five Negative. 10. Negative. 15. Negative. 20 Negative. 25 Negative. 30. I think that will cover everything. Eso for the 1st 0.4 seconds. So 0.4 seconds would be out too. Here. I mean this in a different color. Out to here I am maintaining 10 m per second. So 10 years per second is here staying right horizontal. Then it changes the negative 30. So drop straight down e do it straight all the way down the negative 30. It stays there from 300.42 point 6.6 Would be, um, the small segment out to here. Yep. So then a negative 10. So it hops up, but still stays on the negative side and it stays at that until one seconds. That's all the way over here, Then it could hops up 2.33 So a little less than five somewhere in here. And it stays at that till 1.6 or six. That would be all the way out to here. So hugging kind of a time access. It helps up to 15. So 15 would be up to here for the remainder of the time, which would go out two seconds. Should be Thio here. He's supposed to be horizontal lines. That's pretty much what the velocity versus time graph should look like, given the position time graph. So this would basically be what the graph should look like.

So you're given a position versus time graph. You have tow plot the velocity versus time graph of you wanna find zero points if you can, I think to start. So any time the graph is transitioning from a, uh, positive sloping, um, tangent to a negative slope. Intention. There's a transition point where the velocity is gonna be zero. You're looking at tangent to the curve for these graphs. So the graph object initially as a positive velocity because attention with curves initially positive. But then at some point, it reaches zero. So it it starts out with some fairly, you know, positive velocity. And it's linear. Oh, I didn't know that very well. We go back so you don't do this with tools. Eso it goes from there straight down at some 0.20 we don't know the exact times. Um, it stays negative then so it transitions looks like the curve is fairly uniforms. So it stays negative until it gets the point where transitions, um, from a, uh slowly increasing negative velocity to necessarily decreasing the velocity back to zero again, back to zero. Looks like it's about two thirds out. So maybe possibly out here somewhere. Let's just pick a point I must say about here. Um does look like the second part of that. A little bit steeper. So faster velocity after it transitions. Looks like it's fairly uniforms going from an increasing velocity decreasing negative velocity. Um, till I get to that transition points that before the transition looks fairly uniforms. So I'm gonna say that is going to continue straight down. Um, let's see. So let's see if we can do this. So it's gonna pivot going from a negative slope that's going to stay in negative slope, but of increasing slope to a decreasing speed or slope. And so I'm going to draw it in like this, continuing the same slope down to here, and then it's gonna head back Thio here. Then it turns into a positive slope, and it seems like a much steeper transition. So, yeah, let's see the losses in the beginning faster at a uniformed rate, Um, after it passes that transition point, but it seems like that's gonna be a little bit steeper. So turns positive after the transition point, and it seemed to get slightly steeper. So probably something like this. So the velocity versus time graph would have an appearance when sketch something like

Yeah. Okay, so we're going to make position time graphs and acceleration time graphs for these three. Right, So let's start with, acceleration is probably easiest. So I'm going to draw acceleration times grafts for these three velocity time graphs. Right? So like that, that like that. Okay, so remember to get acceleration from a velocity you're looking for the slope the slope of velocity time graph using the acceleration. So see how here this is a positive slope. Right? So that means this is going to be a positive acceleration. And so since we don't have numbers, I'm just gonna draw it. I'll draw this one in do acceleration green, so it's gonna be some positive value and it's a constant slope. Right? So this is going to be a flat line to represent that the acceleration is some value over here. Okay, so for this one, same thing. This first part is a constant acceleration, right? That velocity is slowing down at first and then it stops and then it speeds up, right? But it's a constant value because the slope is constant. So up until about this point right here, right, the acceleration is constant and it's negative because that's a negative slope. So it's going to be maybe like this. Alright, so again, flatline and then here the acceleration is zero because you're traveling a constant velocity. So this line will jump up and look like that. So yes, I understand like that doesn't really happen in real life, but I'm just drawing it. So it doesn't look like it's discontinuous. Right? So there's your second one and then your third one there's three parts right? So I need to split it into that part, that part. And the in part. Right? So, again, first part slope is zero slope is acceleration, right? So no acceleration because you're going to constant velocity. Second part has a positive slope, so that means you're going to have a positive acceleration. And then the 3rd part is again, no slope means no acceleration. So those are my three acceleration time graphs. All right, so let's draw position time graphs. Right? So this is position time, position time and to get position time. Okay, so I'll do these in blue. So here I have something that's accelerating at a constant rate, right? But This represents zero m/s. So this is something that for this first part that gets slow and then it gets fast. Oops, so I have something that slows down to a stop and speeds up the other way and it starts with the negative velocity and then is positive. Right? So if these are big numbers up here and these are little numbers here, this is something that's starting fast, slows down to a stop and then speeds up the other direction. Alright, we're at this point right here, Matches up with that point right there with the velocity zero there. So the second one it looks kind of like the first one, right? Except it started positive. So it was going in the positive direction. Right? Remember positive is going to be that way this one went down because negative was pointing down, right, so this one start with the positive velocity, slow down, stop here, right, so right there is where I need to stop and then it's going to speed up the other way, but then go a constant speak. So I would start here with a fast positive velocity slow down, stop right here, right, and then I'm going to speed up the other way two right there, right, and then you go a constant speed. So at this part you're going to constant speed in the negative direction. So this kind of looks bad and ideally you would want the slope right here to be the same as the slope here, but this part is a constant velocity and this part is curved, so it's something like that. All right, so you can check yourself to this is a negative slope here. Remember the same thing here, slope of position, we'll give you access velocity so the same thing. Okay, and then let's look at this one. So I have a constant negative velocity. So I start with a big negative velocity and its constant so I'm going backwards so it's gonna maybe go like this for that first part. Right? And then here it's gonna slow down. So slow down. Stop speed up the other way. Rights at this part is a parabola and then here it's going to go a constant speed in the positive direction, so it's going to take whatever slope I had there and go up. Alright, so I've got these two parts, so these these two are constant velocity and then I've got a parabola right there, that's what those grass look like.

In this question were given two stones are initially thrown in the upward direction with speed. The first stone is thrown with the speed of uh u one equals 15 m per second. The second stone is being thrown against Hamilton's with the first in the operation, but with a different speed, which is equal to 30 m per second. The each other stones were initially over a cliff which was at the height of 200 m. The question is we have to verify the displacement of the position of the second stone with this beautiful stone during the motion as represented by this graph. Uh And in the second part, we have to write down the equation of these straight and the cold part of the graph. So for stone, one for don't One or for the first time you can write that, its initial speed is 15 meter per second. If we assume the operation of the positive extraction in that because the acceleration of the of the stories are always going to be minus G, which you can take it as minus 10 m per second square. So as the stone is initially at a height of 200 m from the ground. So then we can write with respect to ground with respect to ground, the displacement of the tune. Ah After a time of the is that so that is defended by X one. So that is going to be equal to uh initial height, which is 200 m plus. You want into t minus half G. T square. So then over here we can write if if after, do you want time? The stone hits the ground, then X one will be called to zero at time T equal to T. One. So follow the uh the equation, we can write zero equals to 200 Plus. You want to go to 50 times T 1 -2. You can take 10 actually the magnet of G. Already the minus and I have taken uh to incorporate the direction of motion into uh do you want to square? So if we simplify this We will get the one square minus three. D. One -40 equals to zero. On simplifying this we can write this as T one minus eight into T. One plus five equals to zero. So from this weekend we'll get the one equals to eight seconds and 151 is Technical -5 seconds. And as the negative value of time is not Permissible. So I can just say that after a time of 8 2nd the stone will be fashionable village into the ground. So as we can see in the graph that at a time of The one equal to eight seconds the first one will be our living with the ground over here. Now, for the second story, we can write similarly for second stone, the initial speed of the second stone is 30 meter per second. The acceleration is again minus G, which is going to be called a minus 10 m/s squared. Uh the initial height is around 200 m. So we can write in this case the displacement of the secretary at any time. T is going to be equal to 200 plus You too, into T -1 G. The squid. This is at any time. T. So ah then on the eve, the time taken that time taken by the second stone is keep to two a rich too ground. Then we can write at The time equal to T. two X. To the displacement of the story. The ground will be zero. So following that, you can write, we can just arrived. The previous equation has zero equals 200 plus You two is 13-32 -1 the we can put a 10 into did to square. So on simplifying this will get ah that T two square minus 6 30 to minus 40 equal to zero. Further simplifying will get t minus 10 into The Plus four equals 0. So from here we'll get the value of T. To the T. Two T two. We will get one as 10 seconds and another 52 will get -4 seconds. And as again the negative well of time is not permissible. So you can say that the second stone will hit the ground after a time of 10. Check it So that we can see from the figure that overhead. The time is 10 seconds when the second story is falling. So then from here we can conclude that for initial eight seconds both uh Stone one and stone to are in motion but for for the time interval of 8 to 10 seconds uh Only stone uh two is in motion and stone is addressed so we can just right for time interval of 02 eight seconds. The stone one and to our in motion, but four time interval of eight, second to 12th, first stone is at arrest and second stone will be in motion. Okay this weekend, right in time interval of 0-8 seconds, the relative motion of stone too, with respect to stone one is Like the relative displacement is going to be X 2 -11. That is going to be equal to X two is 200 plus 30. The -1. The discredit is going to be half G h turn into t square minus again. 200 And this is going to be -15. The class half into turn into the square. So if I simplify this I'll get X two minus x one S 15 T. So this equation is only valid in the time interval of uh zero to eight second. Okay in this time interval you can see that extra minutes X one is proportional to T. Which implied that extra minute explanation will be uh having a linear evolution with T. So this is the Graph that is given in this time interval of 0-8 seconds. So this I can try that extra -11 equal to 15 T. This is a question for the straight line part. Now, after the time interval of eight seconds. So we can say after are in time interval in interval of eight two 10 seconds X one is going to be called to zero and so extra will be Extra equals 200 plus 38 e minus half in two G is turning to the square. So uh Following this weekend right then X 2 -11 is going to be equal to only this 200 plus 30 teeth -5 T square. And as we can see, so remember, this question is only true in the time interval of the less than 10 seconds and more than eight seconds in this time interval. So following that, I can say that over here, X two extra minutes X one is proportional to the square which is a parable expedition. So this will give a parable exploration. So as you can see in the graph uh for this portion, Uh in the diamond level of 8-10 seconds, the graph is a parabolic graph, as you can see uh in the graph. So the equation of this court part of the graph can be very generous, 200 plus 30 t minus five The square. So this is the equation of the cold part of the graph. So this is the answer to the question.


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