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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 $t ?$

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

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 5

Other Trigonometric Forms

Integrals

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Lectures

03:09

In mathematics, precalculu…

31:55

In mathematics, a function…

01:40

Find the velocity and acce…

02:09

Acceleration to position G…

02:58

Find the position function…

Starting from rest, a part…

04:43

Find the velocity, acceler…

01:59

03:52

01:23

00:52

If the position function o…

01:17

So for this problem, you want to find the velocity and acceleration of our particle With this position function, I'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'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're going to need to use the chain will here. So we'll start with the derivative of sign That's going to be the co sign of to t. Oh, and then we'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'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'm so our first term is again going to be the chain rule. I'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.

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