solve-the-given-problems-by-integration-the-acceleration-a-in-mathrmf-mathrms2-of-an-object-is-asin-

Question

Solve the given problems by integration.
The acceleration $a$ (in $\mathrm{f} / \mathrm{s}^{2}$ ) of an object is $a=\sin ^{2} t \cos t$. If the object starts at the origin with a velocity of $6 \mathrm{ft} / \mathrm{s}$, what is its position at time $t ?$

Taylor Shimono · Accepted 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.

Amit Srivastava · Answer

we are doing 80 is equal. Do two years to the power minus T by six. Velocity of zero is one in position as they drove off. The function is so we need to find the functions over lost e and the positions. The velocity function is given by integral 80 80. It is equal during the girl off to use toe dynasty divide best six day T It is a wanted to easiest super minus T by six, divided by minus one divided by six This even. So this comes out to B minus 12 in two yrs two for minus daily, right by six less even being ordered blows at your dealer was equal to one. So putting the very often and you know we get minus 12 leaders tow bar zero plus You see, one is able to one So the pharmacy one is able to 30 on my velocity function is minus 12 is to pull minor city by six less 30. So now you have to find the position function in general bridges in general, be the GT. It doesn&#x27;t pulled off in general off minus 12 elitist one minus day by six I was 13 day take. So this is equal to minus 12 into easiest about minus table 60 whereby minus bind by six clustered in D. C. So this is equal to 72 years to the bottom Dynasty by six Lester DVD. Let&#x27;s see the winner without additional zero is a good reward. Zero to put in the early off the year zero I had is equal to zero, so therefore messy do is equal to minus 72. The final was station function is equal to 72 years to four miners by six plus 13 d minus their vindictive.

Ernest Castorena · Answer

the acceleration function is equal to is T sign of teeth. The initial velocity is given by two negative six. The initial position is given by 10 4 Well, we confined our velocity by taking the anti derivative of acceleration, the respect of time. We&#x27;ll get some unknown constant that we gets all for with our initial condition, wandered with t squared over two in this time and then a negative co sign of teeth in this term, plus an unknown constant record. See, let&#x27;s see. Looks good. Now we can plug in zero. Well, what happened? Swimming plug in zero. We end up with zero on our first component here, and co sign of zero is just one. But there&#x27;s negative in France, we end up with a negative one, and then, plus are unknown elector or give us to negative six. Well, that&#x27;s all for see. We can add the specter to the other side and end up with two negative five. Once we had So our velocity function. We can actually just add this specter and what we solve for C component ways. Teoh. Let&#x27;s see t square over to Plus Two and negative co sign of teeth mine right And we can take the anti derivative one more time when searching for our position function on TV for the position. The 18 we&#x27;ll end of west. Let&#x27;s see T Q over six last duty and on the side of a grand that with negative sign anything minus five t and then we can&#x27;t forget about our new constant director that we need to add. And so now let&#x27;s make use of our initial information on my blood and zero. But you this component becomes zero on the second component becomes zero to you this time. So we just have plus deep is equal. Teoh 10 and four military Well, I just says that are constant. Rector is 10 for So to find our position function, we can add these two vectors component ways are constant and what we integrated to get. So we&#x27;ll end up with seeing cute over six plus two Team Woz 10. And in the second component, we&#x27;re gonna end up with negative sign guilty minus five of deep plus four. So you get awesome

Zulfiqar Ali · Answer

Well, we know the acceleration is equal to D be divided by, uh, DT and we are given x relation. Well, X relation is equal to duty minus six. Now, uh, these equipment can be returnees. D V is equal to ex relation times DT and further D V is equal to okay to t minus six times DT, Right, Okay. And out. Let&#x27;s integrate both sides Then we have integral Well, Devi and the limits are from zero to V and this integral is equal to integral off to t minus six From here the limits on from zero to t and solving integral we have we minus zero physical too D squared minus 60 minus zero. And we also know it. We is equal to D s divided by DT on DST Lottie Petitti u and be different shades displacement With respect to time, we get the lost e right So further disc in the returnees, B s is able to we times DT right and not taken integral on both sides Uh integral mots physical to interior loaf We times gt well here How is do you square minus 60 times TT lives right down limits problems either to s in here. The limits on from zero to t and we have a we is it cool to? Ah, six square minus six into six, which is equal to zero meters per second. Right, So we are looking at Time T is equal to six. We is equal to zero and s distance. Is it cool to 11. 11 to the power three divided by three, minus three into 11. Old square minus zero. And this imply State s is equal to 80.7 meter.

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

Problem

Solve the given problems by integration. Find the…

Problem 45 Easy Difficulty

Solve the given problems by integration.
The acceleration $a$ (in $\mathrm{f} / \mathrm{s}^{2}$ ) of an object is $a=\sin ^{2} t \cos t$. If the object starts at the origin with a velocity of $6 \mathrm{ft} / \mathrm{s}$, what is its position at time $t ?$

Answer

$s=6 t-\frac{1}{9} \sin ^{2} t \cos t-\frac{2}{9} \cos t+\frac{2}{9}$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Anna Marie V.

Kayleah T.

Kristen K.

Samuel H.

Precalculus Review - Intro

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$s=6 t-\frac{1}{9} \sin ^{2} t \cos t-\frac{2}{9} \cos t+\frac{2}{9}$

Basic Technical Mathematics with Calculus

Anna Marie V.

Kayleah T.

Kristen K.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Anna Marie V.

Kayleah T.

Kristen K.

Samuel H.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus