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If the acceleration vector of a particle in motion on 0 plane is a() =< !sin/.sin/ > and thc velocity vector a / 0 is V(O)Fk 0:4> hentgz)isJ33 > MEEFW ...

Question

If the acceleration vector of a particle in motion on 0 plane is a() =< !sin/.sin/ > and thc velocity vector a / 0 is V(O)Fk 0:4> hentgz)isJ33 > MEEFW 1 KOM K #2,2 2 > V [email protected]

If the acceleration vector of a particle in motion on 0 plane is a() =< !sin/.sin/ > and thc velocity vector a / 0 is V(O)Fk 0:4> hentgz)is J33 > MEEFW 1 KOM K #2,2 2 > V [email protected]



Answers

A particle travels in the plane with position vector $\mathbf{r}(t) .$ Find $(\mathbf{a})$ the velocity vector $\mathbf{v}(t)$ and $(\mathbf{b})$ the acceleration vector $a(t)$ $\mathbf{r}(t)=\left\langle t^{2}+\sin 2 t, t^{2}-\cos 2 t\right\rangle$

So in this video we're given the acceleration, which is one of my hat gloves to J hat and were given the initial velocity and the initial position and were asked to determine the velocity function. Yes, so that the lost city folks should have the position. So first of all, we know that the velocity is the integral of the X celebration. So and were given that the acceleration is I had was to her I must do jihad. So we're going to need to grow that. So the integral of one GT is just t the integral off to this to t do you have? And then we have to add a C because this isn't indefinite. Well, we're given a useful piece of information so that we can determine this. See, we're told that the of zero is one k. So instead of TV, put zero. So zero I have lost to time zero j have plus C needs to equal to one. So this is zero. This is zero. So we're just left with C is equal to one k, huh? All right. Great. So we could take this on pulling you back in. So we pluck that in for See, we get that dft this t I have austerity jihad. So that's our for lost. All right, now the position is just the integral off the velocity. So the integral of tea I hat was to t j a happ plus one k had tea the integral of tea with respect cities just t squared, divided by two integral of two. Tea is just Tea Square and the integral off warden It's just And then, of course, plus, C because this is indefinite were also given that are of zero is one I have so free plug in zero for see we get zero squared divided for plug in zero for tm Sorry, we get zero squared divided by two I have was Nero Square J a happ zero k half see needs to equal to one I had so c is equal to one So now we take this see and we look it back in And if we rearrange, we get he's squared divided by two plus one My hat Dusty square jihad plus TK hat And this is our physician

In this question we have the position that the particle that is Party is equals two P. To the power minus t. Coma. Into the power minus T. We are required to find development director and the exhibition vector for this position vector. So let's see how this whole discussion. We know that. The formula to calculate develop city vector is given by first derivative or the by duty of position vector. Artie now substitute all the values so we get but all sort of activity equals two D. By duty of P. U. To the power minus T. Coma into the power minus T. So this can also be written as be by DT of bu to the power minus T. Coma day by duty of E. To the power minus T. We know that The formula to calculate the differentiation of two functions For the multiplication of two functions can the readiness day by dx of you and T. V. It recalls to we differentiation of you plus you divide the excessive we and we know that the differentiation of you do the power X. It recalls to into the power X. So on the basis of these formulas we can write develop city vector VT records too E. To the power minus T multiplied by the differentiation of tea with respect to T. And this will be called to one plus T. Multiplied by the differentiation of into the power minus T. And this will be equals two minus into the power minus t. Coma. The differentiation of it. Department city will be called to a to the power minus T. So when we for us all this we get velocity vector beauty is equals two. Into the power minus t minus t T. To the power minus t. Coma into the power minus T. So this is the final answer for the part A. And now let's move to part B. Fire to be the formula to calculate the acceleration vector of the particle 80 is given by D by duty of velocity of activity. Now substitute the value of also director. So we get exploration vector 80 is calls to D. By duty of into the power minus t minus T. E. To the power minus t coma into the power minus T. So this will be calls to D by duty of here to give our minus t minus t. Into the power minus t. Coma. D. By duty of minus U. To the power minus t. The venue for the calculate this. Finally we get Exploration, Victor 80 is equals two -2. Into the power minus t plus t. To the power minus t comma here to the power minus T. So this is the final answer for this problem. I hope you understand the solution. Thank you

Okay notice the position vector. Given the first thing we're gonna do is we're going to find the velocity vector by taking the derivative of each of the components. So V. Of T equals three T squared in the I. Direction in one half in the J direction. The next thing we'll do is we'll take the derivative of V. Of T. To get a job. So now we have 60 in the I. Direction and we do not have a J direction. So it's zero DJ. So we don't right now the speed is the absolute or is the magnitude of velocity? So we're taking that three T squared and squaring it and adding it to a half squared. So we get the square root of nine T. To the fourth plus 1/4. Okay, so um we can see that we have t to the third component in the X. Direction. In a linear component in the J direction. So I'm going to be using both negative and positive T. So I can sketch this out appropriately. My negative two is being put into my eye component and it gives me a negative eight. Now fill in the rest the same way to get my wife component. I'll be placing my T. Values into the J. My Y coordinate into the J. Component. Okay so I'll mark off my X. Axis with important numbers and then I'll go ahead and um graph those. So we knew we'd have curvature because of that um t to the third. Um And uh so you know our graph does look like we're exes t to the third and why is linear? So now we're going to look at the value of the velocity at T. Equals zero. If I put a zero in I get zero in the eye but a half and a J. So it ends up that my vector should go straight up to half. Then I can go ahead and put zero into my acceleration and I do get zero. And so um it won't be graphing that, but look at the graph and look at curvature right. Our second derivative is all about our cavity, and Arkan cavity does change at that place. So it's not surprising that our acceleration is zero.

Okay, first we are going to find the velocity, acceleration and speed of our given position vector are empty. So to find Vm T will be taking the derivative of that. So we will get a two T. In the I direction and then a negative one. So we can just say negative in the J direction. R a f T is our next derivative where we get to in the I direction. So our speed ends up being the magnitude of velocity. So we can take the square root of the two T squared plus the one negative one squared. So we end up with a square root of four T squared plus one. Okay, so now we are going to go ahead and graph this. I do have a quadratic in the X direction and a linear and the Y. So it makes me think that I'm going to still have a quadratic shape, but it's actually going to be opening to the right. But let's go ahead and sketch these out. So we can really find out what our real values are because those plus ones also will be moving our vertex. So what I'm doing is I'm just coming up with T. Values and then I'm placing those T values into my eye component and writing those as X values and then into my why component or my J. Component and writing those as why. So now we can go ahead and start sketching some of these. I'm trying to see where on my graphs and my important places are and so now I can sketch my 53, my 22 11 and so on and so forth. And so I end up with a sketch that looks like this and goes in this direction. So um from this point we are going to now look at V. Of zero. So if I did put a zero in notice that it would be zero in the eye and then minus J. So I'm gonna go ahead and graph that negative J. So it's like one unit directly in the negative Y. Direction. And then for my acceleration, if I try to put a zero in, I get eight to I. So that will be two units to the right for our acceleration at time equals zero.


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