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Question 1ed particle' position vectoras function cftime ziven by F) = t"i+2t*j+tk plue particle position Vector?s functicn of time given by F(t) = 2t&quo...

Question

Question 1ed particle' position vectoras function cftime ziven by F) = t"i+2t*j+tk plue particle position Vector?s functicn of time given by F(t) = 2t"1 _ + j++kAtt= 4what the velocity cf blue particl relative tothe red particle? Wnat Ihe red particle' accelerationaft= 25? Is the red particl acceleration perpendiculartoitvelocity a-t= 25? Find unitvector in tne direction cfthe red particle $ accel ration att= 2s

Question 1 ed particle' position vectoras function cftime ziven by F) = t"i+2t*j+tk plue particle position Vector?s functicn of time given by F(t) = 2t"1 _ + j++k Att= 4what the velocity cf blue particl relative tothe red particle? Wnat Ihe red particle' accelerationaft= 25? Is the red particl acceleration perpendiculartoitvelocity a-t= 25? Find unitvector in tne direction cfthe red particle $ accel ration att= 2s



Answers

Find the velocity, acceleration, and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleration vectors for the specified value of $ t $.

$ r(t) = \langle t^2, \frac{1}{t^2} \rangle $ , $ t = 1 $

Hi, everyone. Tim John's here, and we're doing another problem dealing with displacement, velocity and acceleration functions. So this problem were first given the position function as s is equal 2.27 t cubed minus 0.65 T squared minus 2.35 T plus 4.4. So you can use a graphing calculator or a graphing website to go ahead and plot this function. Um, what you should see is that it will look something like this, and we're just doing this for the first five seconds and SSN feet per second and time is in seconds. So whenever time of zero, the position will start at 4.4 right here. And the question is not asking for wherever it crosses zero. So we can go ahead and put that to the side just for now. The next thing to Dio is go ahead and software the velocity, Which is this? The time derivative of the position function here. And this will go ahead and give us 0.8 won t squared minus 1.3 T minus 2.35 and velocities and feet per second. Time is, of course, still in seconds, and this is just a quadratic function. Whenever time is zero, we'll see that it's 0.235 here and it will cross zero and look something like this. We'll do this just for the first five seconds and finally plotting the acceleration. So acceleration is just the derivative of the velocity. So if we go ahead and do that, we'll get 1.62 t minus 1.30 and the derivative of a quadratic will just give us a straight line. So every time is zero. It will start here, and it will just look like a straight line crossing through the T axis right here. Acceleration feet per second squared. Time is in seconds. Yeah, and the second part of the problem asks us what is the positive time when the particle changes? It's direction. So this would be a thing would be at the exact point where the velocity is zero because the velocity would be negative. And then it would switch over to positive. And this would be a positive time that the particle will change its direction. So what we do need to do here is we need to set our velocity function here, set to zero, and then we need to solve for whichever positive time it is. So have 0.8 won t squared, minus 1.3 T minus 2.35 Okay, remember, the quadratic function is minus B plus or minus B squared, minus four A C square root and moving to a. So I have positive 1.3 plus or minus minus 1.3 squared minus four times 40.81 times minus 2.35 square root of that. And then we'll have to times 0.81 on the bottom. So go ahead and solve that. And then you should get that. The two times will get his 2.69 seconds and a minus 1.8 seconds. But we need the positive time. So we're just gonna look at 2.69 seconds. And that is the positive time that the particle changes. It's direction or the time that the velocity goes from negative to positive.

Okay. This question gives us a position function, and it wants us to calculate the velocity and acceleration vectors. AT T equals two. So first, let's graft this particle, and I personally just used as miles for this. So if we do that, you'll see that our particle's path will look something like this. So just like a problem opening that way and then to see what point we care about are vectors that we care about t equals two. So AT T equals two our position vector plugging in to becomes negative to two. So will say that the point we care about at Sequels too, is that blue dot right there. So now let's go about calculating the velocity and acceleration at that point. So for velocity, we know that it's the derivative of position. So let's take the derivative of each component. So we get negative to t over too comma one or that means that our velocity is negative. T one so no, that we know that V of tea is equal. The negative t comma one. We can calculate acceleration, which is just the derivative of velocity and doing it component wise, we get negative one comma zero. So now listing all three of our functions. Here we see that position is negative. 1/2 a T squared comma. Two velocity. Yes. Negative, t one and acceleration is negative. One zero. So now at T equals two for each of these are of two is negative to two V of two iss. Negative to one and a of two is negative. 10 So now we can plot all these vectors together. So our velocity vector at this point is negative too common One so negative to one kind of looks like this. So that's our velocity vector. And then for acceleration Vector, it's negative 10 So our acceleration vector is just a straight line in that direction. So green is V and blue, his acceleration, and there we have it.

I guess we have our up t. It's equal to B squared minus one comma t were asked to find in philosophy. But that's this derivative of Artie. So that's two t common one. Okay, acceleration is this through the derivative of our velocity? So that's just to combat zone and in speed is equal to the magnitude of our velocity, which is square roots. Both 40 squared plus one.

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.


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