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Particle'$ acceleration given by the function .2t' +5t2, find its exact velocity at t = secondsAnswer:Find the exact position the particle from problem wh...

Question

Particle'$ acceleration given by the function .2t' +5t2, find its exact velocity at t = secondsAnswer:Find the exact position the particle from problem when t = secondsAnswer:

Particle'$ acceleration given by the function . 2t' +5t2, find its exact velocity at t = seconds Answer: Find the exact position the particle from problem when t = seconds Answer:



Answers

Find the velocity, acceleration, and speed of a particle with the given position function.
$$
\mathbf{r}(t)=\left\langle t^{2}-1, t\right\rangle
$$

Okay, So this question wants us to find the velocity, acceleration and speed of a particle whose path is described by this position function. So to find our velocity, we just take the derivative of the position function with respect to time. And we do that by each component so derivative of to co sign T is negative to scientist e derivative of three t is three and the derivative of two scientists. E is to co sign t then for acceleration Vector. We just take the derivative of velocity and this becomes negative to co sign t zero and negative to Scientist E. So now we found our velocity and our acceleration. So now the last thing has defined the speed which is equal to the magnitude of the velocity vector. So we know that the magnitude of e of t is just the sum of all the squared velocity components. And since we found our velocity vector already, we can just plug these in So our X component of velocity is negative to sci Fi are y component of velocity is three and r z component of velocity is to co sign to you and remember the square each component. So now doing all the squares here we get for sine squared T plus nine plus for co sign squared T. And now if we do some rearranging, I hope you can see some simplifications that we could make. Because if we factor out before from the 1st 2 terms, they have him in common. We see we get a sine squared t plus a co sign squared T and remember from our Pythagorean identity. But sine squared plus co sign squared T is just one. So this means that our velocity becomes four times one plus nine or squared of 13 is the magnitude of our velocity, which is pretty interesting. Does this?

Okay, so this question gives us the position function of a particle, and it wants us to find expressions for the velocity, acceleration and speed off the particle. So for velocity, we just need to take a derivative of the position. And remember, we do this component by component. So the derivative of the first component is to t plus farm. The derivative of the second component is to t minus one, and the derivative of the third component is three t squared. So that's our expression for velocity and then for acceleration, we just need to take another derivative. So the derivative of the X component of velocity is too. The derivative of the white component of velocity is too, and the derivative for the Z component of velocity is 60. So now we have expressions for both our velocity and acceleration. But we're not done yet because we also need the speed. So speed is equal to the magnitude of velocity. So we're just gonna have to find the magnitude of our velocity function. So the magnitude of e of t is equal to Well, we just use our version of the Pythagorean theorem. So again, that's just t x t t squared plus d Y t t squared plus d C D t squared And we know each of these components from the last part So give us a big square root to work with here. So our X component is to t plus one are y component is to t minus one and r C component is three t squared. So if we do that, what's foil these out and see what we can do Simplifying these so we get for t squared plus 40 plus one plus 40 squared minus 40 plus one men plus nine t to the fourth and now we can see one quick simplification. This cancels with this. So now we get the square root of eight t squared plus two plus nine t to the fourth or registry Raying this in a more standard form the square root of nine T to the fourth plus eight t squared plus two. And this is our final answer

Given velocity function V f t which is equal to t squared minus two t minus three. We want to find the displacement and the distance traveled by the particle Over the interval 2- four. For the first part we have s this is equal to The integral from 2 to 4 of VFT which is just The integral from 2 to 4 of He squared -2 T -3 Bt. And integrating, we have 1 3rd TQ two times t squared over two minus three T. This evaluated from 2 to 4 and simplifying, we have 1 30 cube minus t squared minus three T. This evaluated from 2 to 4 and when these four we have 1/3 times four. Race to the third power minus four squared minus three times four. This minus 20 is too, we have one third times two to the third, power minus two squared minus three times two. This gives us the value of to over three and so the displacement of the particle Is to 3rd m for part B B house distance. This is equal to the Integral from 2 to 4 of the absolute value of VF t d T. And this is equal to the integral from 2 to 4 of the absolute value of t squared minus two t minus three Bt. Now the absolute value of t squared minus two t minus three. This is equal to the absolute value of t minus three Times T Plus one. This is equal to positive t minus three Time's D Plus one. If t is in The interval 3-4 And you have negative of T -3 time's D plus one yes T is in 2- three. And so we rewrite this into the Integral from 2 to 3 of the negative of t squared minus two t minus three D. T. Plus we have The integral from 3 to 4 of t squared minus two t minus three DT. Now integrating we have negative of one third t cube minus we have to t squared over two, that's D squared minus three T. This evaluated from 2 to 3 plus We have 1/3 t to the third power minus t squared minus three. T. This evaluated from 3 to 4. Now when these three we have negative of one third times three to the third power minus three squared minus three times three. This -20 is too you have one third times two to the third, power minus two Squared -3 times two. And then this plus we have when these four that's 1/3 Times 4 to the 3rd. Power minus four squared minus three times four. This minus when T is three we have 1/3 times 3 to the third power minus three squared minus. We have three times three and this gives us Negative of we have 9 -9 minus nine minus. We have 8/3 -4 -6 and then plus we have 64/3 -16 -12 minus. We have 9 -9 -9, and so this gives us the value of to over three, and so the Distance traveled by the particle is 2/3 m.

Okay. This question wants us to find the velocity, acceleration and speed of a particle with this position function. So to find the velocity, we just need to take the derivative of the position function which doing this component wise Rx component derivative, is to t our white component derivative is co sign t minus. Then we have to do the product rule co sign T minus T sai Inti and then for 1/3 component, we get negative scientist e plus scientist e plus t CO sign t and then we can see some simplifications. So we see that V A T just simplifies to Because this cancels and this cancels so just simplifies to to t in our first component, T scientist Ian, are y component antico sign t in our Z component. So from here we confined our acceleration, which is the derivative of velocity And then again, going component list We see our derivative of our first term is too. Our second component is Sai Inti plus Co sign Sorry t co sign T and the derivative of our third term is co sign T minus t Sign t So now we have a velocity and acceleration. So now we can just find our speed, which is the magnitude of velocity is the square root of the sums of the squares of the components. And we can just plug these in so our X component for velocity is too t are y component for velocity is t scientist E and R Z component is Tico santi. So now it's just square all these can we get four t squared plus t squared sine squared, t plus t squared co sign squared to you, and you can pull a couple things out here. So first we can pull a t squared out to give us sine squared t plus co sign squared T Plus four. Or we can usurp a faggot and identity to get T squared Times one plus four or square roots of five T squared, which simplifies two t Route five as our magnitude So T Route five is our philosophy, and we found our acceleration and velocity earlier


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