find-the-velocity-and-acceleration-of-a-particle-whose-position-function-is-xtt3-9-t224-t-t0

Name: find the velocity and acceleration of a particle whose position function is xtt3 9 t224 t t0
Uploaded: 2019-08-25
Duration: 1 min 23 s
Channel: Numerade
Description: Find the velocity and acceleration of a particle whose position function is $x(t)=t^{3}-9 t^{2}+24 t, t>0$

Question

Find the velocity and acceleration of a particle whose position function is $x(t)=t^{3}-9 t^{2}+24 t, t>0$

Taylor Shimono · Accepted Answer

So for this problem, we can go ahead and start by finding the velocity of our particle. I also know that the velocity is just the derivative of our position function. So to find our velocity, all we need to do is take the derivative of T cute. So we&#x27;ll move this three to the front and reduce the export of my one. So this is three t squared, minus the derivative of 90 square. Again, we&#x27;re going to move this tea to the front. So two times nine is 18 times t to the first power and finally plus this 2014. Essentially, we&#x27;re just going to be the one to the front and we just have plus 24. I&#x27;m So now that we have our velocity function, we can find our acceleration function the same way. So we know that since the velocity function is the derivative of the position function, the acceleration function is the second derivative of the position or the first derivative of the velocity. I&#x27;m so essentially we know this is the acceleration function, and all we have to do is differentiate this philosophy function. I&#x27;m so the derivative of three t squared again. We&#x27;ll move that to to the front and reduce our power by one. So that&#x27;s 16 minus. Since we have just a t on the 18 this is just going to be minus 18. And finally, our constant 24 is going to leave us with plus zero at the end. So here we have our velocity function in our acceleration from

Foster Wisusik · Answer

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&#x27;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&#x27;re not done yet because we also need the speed. So speed is equal to the magnitude of velocity. So we&#x27;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&#x27;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&#x27;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

Amy Jiang · Answer

I guess we have our up t. It&#x27;s equal to B squared minus one comma t were asked to find in philosophy. But that&#x27;s this derivative of Artie. So that&#x27;s two t common one. Okay, acceleration is this through the derivative of our velocity? So that&#x27;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.

Taylor Shimono · Answer

So for this problem, you want to find the velocity and acceleration of our particle With this position function, I&#x27;m so right away. We know that the velocity function of e of teeth is actually just the first derivative of our position function. I&#x27;m so all we need to do to find our velocity is take the first derivative of this except T s. So the derivative of sign two teeth, we know that we&#x27;re going to need to use the chain will here. So we&#x27;ll start with the derivative of sign That&#x27;s going to be the co sign of to t. Oh, and then we&#x27;re going to multiply that by the derivative of the inside. So times the derivative of two t, which is dressed too and then finally plus the derivative of coastline teeth will end up being a minus sign T. I mean, so I&#x27;m going to go. And simple far this to our velocity function is equal to two coastline duty minus sign of teeth. And lastly, we want to find our acceleration function. Actually, no, the acceleration function is actually just the first derivative of velocity. So again, all we need to do is take the first derivative of this function that we found right here. I&#x27;m so our first term is again going to be the chain rule. I&#x27;m so are constant will not change anything Times the derivative of co sign to t I was going to be minus sign in to tea times again the derivative of the inside portion here. So the derivative of two t, which is again just to and then finally minus the derivative of scientific is just minus co sign teeth. And once again we simplify. We find that our acceleration function is equal to negative for sign to t minus co sign t.

Foster Wisusik · Answer

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&#x27;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

Taylor Shimono · Answer

So for this problem, we want to you find our velocity and acceleration functions of our particle. But before we do that, we can see that we have a trig identities identity here in our position function, um, our salary and identity, which, if we factor out to here, we get two times sine squared t plus co sign squared teeth. And we know that signs where t Plus co sounds great. Tea is just equal to one. So our position is always going to be at two times one, which is just too, so we can see here that this is a constant. So when we find our velocity, we know that we&#x27;re taking the derivative of our position function. So we&#x27;re just taking the derivative of a constant which is going to end up being zero and the same thing for our acceleration, which is the derivative of velocity. So again, the derivative of zero is just going to be zero. So our velocity and acceleration

Problem

F. NO CHANGE G. tells us, that, H. tells us that,…

Problem 1 Medium Difficulty

Find the velocity and acceleration of a particle whose position function is $x(t)=t^{3}-9 t^{2}+24 t, t>0$

Answer

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