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Aparticle moves In the xY plane with a constant acceleratlon glven by a=-4 Jm/s? At t = 0,Its position and velocity are 9 Im and (-21+8 ] ) m/s , respectively: What...

Question

Aparticle moves In the xY plane with a constant acceleratlon glven by a=-4 Jm/s? At t = 0,Its position and velocity are 9 Im and (-21+8 ] ) m/s , respectively: What Is the distance (In meters) from the orlgln to the particle att= 2.0 s?

Aparticle moves In the xY plane with a constant acceleratlon glven by a=-4 Jm/s? At t = 0,Its position and velocity are 9 Im and (-21+8 ] ) m/s , respectively: What Is the distance (In meters) from the orlgln to the particle att= 2.0 s?



Answers

A particle starts from the origin at $t=0$ with a velocity of $8.0 \hat{\mathrm{j}}+2.0 \hat{\mathrm{j}}$ moves in the $x y$ plane with constant acceleration
$(4.0 \hat{\mathrm{i}}+2.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2}$ . When the particle's $x$ coordinate is $29 \mathrm{m},$ what
are its (a) $y$ coordinate and (b) speed?

Hi there. So for this problem we have a particle with the given conditions at the equals to zero seconds. The velocity is eight m per second only in the white component and it also moved with a constant acceleration that is also given for meters per second square in the X company and two m per second square in the why company? Now what we need to determine for the first part of this problem part A is mm The white company of its position when the ads component or x coordinate of this particle is 29 m. Now in order to do that we can use the backdoor notation or we can use the the coordinates separately. The ads company and the white component. Now we are going to use that. So the X component is equal to the initial sp the the X company of initial speed times the time plus one half of the acceleration times the times squared. And for the white company we have something similar. We have that that is the initial competent of the speed times the time plus one half of the acceleration, times the times square. Now for the first for the X company we will have that. We don't have in here any X. Component. So the at the company of that initial velocity is just zero. So we don't have the first term. The first term is zero and now and here we have the X company of the acceleration and in the other one the white company. But we indeed have uh a next company of the acceleration. So we just substitute that in dirt and sold for the time. And then when we obtain the time, we can substitute it in the other one to obtain the Y coordinate. So that's what we are going to do. We know that um solving for tea we will have that that is two times EDS over the X company of the acceleration. The square root of this will give us the time. So we just substitute those values. We know that the Adds value. The x coordinate is 29 m. The X component of the acceleration is four for meters per second square. So from this we obtained at the time, Yes 3.8 seconds. So with this we can substitute in the second equation for the white component or the y coordinate and we obtain the following, we need the initial component of and the white company um The white component of the initial speed which is eight. So we put that in here eight m per seconds. Heinz the times that we are obtained 3.8 seconds plus one half of the white company of the better acceleration which in this case you can see that is to so we have two m per second squared and the time that we obtain To the square. So putting all this into the calculator, we obtain that the white coordinator of this particle is 45 m. So this is the solution for the first part of this problem. Now for part b, what we need to obtain it's the speed of this party gold. This speed is the magnitude of the the velocity. Now we know that the velocity of the particle is given by the following the initial Velocity plus the veteran acceleration times The times it has passed where we know that the time is the one that we obtain before 3.8 seconds because we want the velocity at that time. So what we need to do is to substitute all of those values. We know that the initial speed is eight m per second in the y component. We substitute the better, the acceleration better which is four m per second squared in the X component plus two m per second. Square in the white competent this times the time which is 3.8 seconds. So solving for this, we know that way we are serving bettors we can only assume the white companies with the white companies and the X company in this case we just only have one. So solving for this, we obtain that that better is 15.2 m/s in the white come in the x company and 15.6 right? Yeah 15.6 m/s in the desk and the white components. So what we need to obtain now is the magnitude of that better. So that is the square root of each component. To square the some of this. Yeah. so from this we obtained, the magnitude of this spectrum Is equal to 22 m/s. So this is the solution for this problem.

In this problem we are given two vectors, one for the velocity and one for the acceleration of a particle. And after some time. But particle is that X equals 29 m. And we want to find the white point which it is when it has traveled those 29 m in the X axis. And we know an equation for displacement. Because what we really are trying to find is a displacement. So let's say that why the displacement in the Y axis equals to the initial velocity. No Y axis times the time it has passed. Mhm. Plus one half times the acceleration in the Y axis, times time square. We have all this information except for time, but we can find time By knowing that the particle has traveled 29 m. So let's do that X. So about a very different car. Yeah. Yeah, X equals to one half times the acceleration in the Y axis times, Times Square. Now we don't have an initial velocity in the Y axis. That is why we don't even bother writing down that term. So we solve for time and we find that time is equal to the square root of two times the displacement. The X axis all over the acceleration in the X axis is we have a we have a type official. All right. And this is equal to two times 29 over the acceleration in the X axis. Which is for so this is going to finally be equal to almost 3.8 seconds. So now we have all we need to replace the information in this equation. So let's do that. And with a calculator we find that this displacement is equal to 44.84 m, which is Roughly 45 m. And that's the answer for part eight. Now let's move on to Part B. But we're trying to find the speed once the particle is at this Y and X locations. Mhm But to find a spear, we need to find the components of the velocity at that point. So let's do about first. We X. Is going to be equal to the acceleration in the X axis times the time. Yeah. And the velocity in the Y axis is going to be equal to the initial velocity in the Y axis plus the acceleration in the Y axis times time. And as we said, we don't have an initial velocity X access. That is why this term for the X Values zero but we do have one for why. So now let's replace the day to begin. We have the acceleration and we have the time we found which is frequently Over here we have eight m/s plus two times the time. Mhm. Mhm. So that is going to give us 15.2 meters per second. And over here that is going to give us 15 .6 m/s. And now we can find the velocity or I mean the speed very easily by finding the square root of the X component squared plus the Y component squared. That's it. Mhm. Mhm. So now let's substitute that. And with a calculator, we find that this is almost 22 meters per cent and that is the finalized.

All right. The particle starts at the origin and has a velocity of eight in the J direction meters per second and its acceleration is four in the I. Direction plus two in the J direction meters per second. Yeah. So I'm trying to figure out mhm. I'm gonna do time first when the X coordinate is 29. So X is gonna equal X. Initial which is zero plus V. X. Initial which is eight times T plus one half times the acceleration, which is four T squared. So X needs to be 29. So um I'm gonna rewrite this as um to T squared plus 80 minus 29 equals zero. And now I'm just going to graph it and look for the Y intercept to T. Where'd that's a T. Minus 29. So I'm graphing it and I see that the X intercept is at 2.30 seconds. Yeah. So I'm gonna write T equals 2.30 in my calculator. Okay, so the question now is in 2.3 seconds. What is its? Why coordinate? Well why is going to equal why initial? Oh no, that eight that eight was in the and the Y. Direction? Uh Yeah, so X is gonna equal Vieques which was zero plus one half 80 squared. Oh man, so that X again that eight is again in the Y direction, so it's one half a T squared. Ok, so now I just have to take the square root of 29 over to Mhm man, T equals square root of 29/2. So that's going to be 3.81 seconds. All right. Now, what's the Y coordinate? Why is V. T um plus one half a. T squared? Those cancel out? So it's just gonna be 80 plus t squared, 80 plus t squared. It's 44 45 40 45 meters. And this is question number 43. So let's see if I'm right. Yes. Now um what's its speed? What is its speed? Well, v final is going to be uh 80 it's gonna be four T. In the X. Direction and then it's gonna be the initial which is eight plus two T. In the J. Direction meters per second. Sell. Put that in a calculator. Fourty is 15 15, eight plus two T. Is 16 meters per second. But let's see what we got. 43. It says 22 meters per second. Well that's the speed. Oh it does ask us for speed. Okay. Square root of 15 squared plus 16 squared would be the speed. I thought it was asking for a velocity. Okay. Not in a calculator gives me 22. Thank you for watching. Mm.

In this problem it has given that any self velocity is 10 Jacob meter per second. So you victories 10 Jacob and acceleration of the particle is eight IPAP plus to Jacob. So eight I kept plus to Jacob. So what will be the value of velocity vector at any instant time? T soapy vector will be caused to you victor plus a vector into t. So you vector here is 10 Jacob And the victories eight icap Plus to Jacob into T. So this is equals two 80 I cap plus 10 Plus two T. Jacob. So this is the expression for velocity vector at any time. T Okay so let us say this is a question one. Now can you find the position vector using this equation? So velocity vector can be written as the rate of change of police in vector. So D R by DT is equals two. Eight T icap list 10 Plus two T. Jacob. Therefore dear victor will be close to 80. So eight DDT. I kept less 10 plus two T. D. T. Jacob. Now we can integrate both sides. So integration of director will be close to integration of a T D. T. I cap plus integration of 10 plus two T DT Jacob. Okay So here uh at equals to zero. His art his are was zero. So at T close to T his position victories. Let us say our victor. So what will be the result of this integration? So our vector will be close to the integration of 80 will be 40 square so 40 square. And if you put the limit of upper limit so parliament will be T square and lower limiting all the cases is going to be zero. So It's need not to write those terms so better to write only the of Parliament's so 40 square plus integration of 10. So integration of 10 with respect to T will be 20. Okay uh Here uh the unit is in icap. These are capped coefficient and now we are going to write for the Jacob coefficient. So 20 plus integration of duty will be to T square by +222 will get cancelled. So the only remaining time will be the square. So plus t square Jacob. Okay, now what is the value of position vectors? So position vector is basically X I cab plus why Jacob? So this will be equals two. 40 square Icap, blessed. 10 T plus t squared Jacob. Okay now in the problem with it saying that the X coordinate of that point is 16. So it means that 40 sq is equal to X. That is equals to 16. So can we find the value of T. So we can compare these two. So if we divide both sides by four so T square will be equals to four. So T will be equals to two. Now if th equals to two then what is on comparing the coefficient of y cap? What we can say why is equals two 10 T Plus T Square. So why is equals two 20 plus T squared. So he had tears too. So it will be 20 plus two is square. that will be caused to 24. So why coordinate of that point is 24. So this is the answer for first part nine, part B. It is saying that what is the magnitude and direction of velocity of the particle at equals two. I'm sorry what is the in be part it is saying that what is the speed of the particular at that time? So we have found the value of T. Which was two. So we will use this value in this in this equation one. So, equation one was 80. Icap, velocity is 80 icap plus 10 plus to T. Jacob. Okay, the value of these two. So putting the value of the coast to to velocity vector will be 16 icap plus 14 Jacob. All right. So now we can find the value of magnitude. So magnitude will be equals two mother of the victim, which is equal to the square root of 16 square plus 14 is squared, which will be caused to 21 point To six m/s. Now, how to find direction. So direction will be equals two. 10 universe 14 divided by 16. So 10 in verse 14 divided by 16 gives 41.18 degree. So velocity is making an angle of 41.18 degree in the counterclockwise direction with the positive x axis. So, these are the answers for part B.


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