5

The velocity of a particle is given by vd)-(2+-1Ji+(2-t)j+(3+-4Jk Find the acceleration of this parlicle a5 a function of time; ( in unit-vector notationSelect one:...

Question

The velocity of a particle is given by vd)-(2+-1Ji+(2-t)j+(3+-4Jk Find the acceleration of this parlicle a5 a function of time; ( in unit-vector notationSelect one: 41j+3kb. 2i-2j+3kc.2i-j+3k2ij-4k

The velocity of a particle is given by vd)-(2+-1Ji+(2-t)j+(3+-4Jk Find the acceleration of this parlicle a5 a function of time; ( in unit-vector notation Select one: 41j+3k b. 2i-2j+3k c.2i-j+3k 2ij-4k



Answers

(III) The acceleration of a particle is given by $a=A \sqrt{t}$
where $A=2.0 \mathrm{m} / \mathrm{s}^{5 / 2}$ . At $t=0, v=7.5 \mathrm{m} / \mathrm{s}$ and $x=0 .$
(a) What is the speed as a function of time? (b) What is the
displacement as a function of time? (c) What are the accel-
eration, speed and displacement at $t=5.0 \mathrm{s} ?$

Okay notice the position vector that we have been given. We are first going to take the velocity of that by taking the derivative of each of the components. So V of t equals four, I minus three T squared J. Now, in order to get the acceleration will go to the next derivative. So the I. Component will go away but we have a negative six T. In the J. Direction. Okay, so um in order to graft so first we're going to find speed. So speed is the magnitude of velocity. So we will do the square root of four squared plus negative nine t squared squared. And so we get that square root of 16 plus 92 the fourth. Okay so now we're going to graph it. So in order to do that we come up with some teas and it's usually good to have both positive and negative teas and then we put the tea into the I. Component and that will give us our X value. So negative two times four will be a negative eight negative 40. And then we'll go back to four and then eight um are Y. Component is put into the negative T. To the third. So t to the third of negative two would be a negative eight but then it goes positive And then it's the next value would actually be a -8. So let's go ahead and sketch some of those numbers out. So kind of using my bearings on my X. Axis. I'm able to kind of quickly figure out what those values are and then knowing that I'm going to um 8 -8 is useful. Okay so now I'm finding V. Of zero so I go put zero in for time and I get four. I plus zero J. So my velocity at zero. That instantaneous velocity is all to the right and it's a magnitude of four. Now my acceleration when I put a. T. Value in I get zero. Now does that make sense? Well remember your second derivative, which acceleration is the second derivative of that position function? Your second derivative has to do with con cavity. So we go from being concave up to convict cave down. So right there are curvature actually is zero. And so that's why we get that value.

Okay first we're going to find the velocity acceleration and speed of our given motion for our particle. So in order to find our velocity, we take our first derivative. So our Vot is just going to equal one eye so we can just write I and then minus three T squared in the J direction. Our acceleration will be the next derivative, no I. Component. But we have a negative six T in the J direction. So now our speed is the magnitude of velocity. So we can go ahead and express that as the square root of one square plus that negative three T squared. So it's a one plus 92 the fourth. So now we're gonna go ahead and graph this so notice that were linear in the X. Direction but we're a degree three in that Y direction. And so with that you know, we could obviously put some points in but what we end up with is a function that this would be a positive X. to the three. But because that is a negative T. To the third, it ends up going this direction. Now when we go to actually um find our values for Visa zero, if you put a zero in notice you get I minus zero J. So because of that I we know our velocity is one unit to the right now. Our acceleration when we put a zero and we just get zero and what you might notice is we go from a positive curvature to a negative curvature for that second derivative. So you can kind of see how it changed. So it's not surprising that our second derivative ends up being zero at that point.

For party to find the velocity vector because a velocity defector function as a function of time, This would be equal to the derivative with respect to time of the position function. So this would be I hat plus four t squared J hat plus T K hat. And this is simply gonna be equal to eight t j hat plus que hat. This would be the function for the velocity equations, velocity function. And for part B, we want to find the acceleration function as, uh, with respect to time. And so this would be equaling the derivative with respect to time of the velocity function. So eight T j hat plus que hat. And this is simply gonna be equaling 28 j hat. The units would, of course, be meters per second squared. This would be our answer for part B. The units here also meters per second, Of course, for per day, that is tthe e end of the solution. Thank you for watching

He It's clear. So a new breed here. So we're gonna find the velocity and acceleration. So we have V A. T is equal to Ford t cubed minus 60 square plus two t plus one. It's the derivative of our loss original function, the position function. And then we're gonna differentiate the velocity to get acceleration. No, becomes 12. T square minus 12 T plus two. That's part A For part B. We have We're gonna use our acceleration that we found in part a. Then we're gonna plug in one. When we get two meters perverts stuck in square for part C, we're going to draw a graph so we have our regular function or position function in black. Then I'll look like this. We have our velocity function and red and our acceleration function gangrene


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