solve-the-given-problems-by-integrationconditions-are-often-such-that-a-force-proportional-to-the-ve

Question

Solve the given problems by integration.Conditions are often such that a force proportional to the velocity tends to retard the motion of an object moving through a resisting medium. Under such conditions, the acceleration of a certain object moving down an inclined plane is given by $20-v$. This leads to the equation $t=\int \frac{d v}{20-v}$. If the object starts from rest, find the expression for the velocity as a function of time.

Jonathon Brumley · Accepted Answer

we are trying to show that the position function comes out the same regardless of the method that it is obtained. First, let&#x27;s look at the anti derivative method, which says that the position function is nothing more than the integral of the velocity. So let&#x27;s integrate our velocity function nine minus T square. The integral of that is nine T minus T cubed over three, plus a constant. Now the position function, it zeroes negative too. So if I fill in zero for the time, I should get a position of negative too. So let&#x27;s see, what see is position functions negative to in time is zero by filling zero for the time that zero plus zero plus c therefore see is negative two. So the position function using the anti derivative method is 90 minus t cubed over three minus two. Okay, now, the second method is using the fundamental theorem of calculus that would tell us that the position function s A. T is equal to the starting position s and zero plus the integral from zero t of the velocity of X dx already established. The position function at zero is negative two and two. That the integral from zero to t of V of X dx. Okay, so the velocity it X would be nine minus x squared integrated. That would be nine x minus X cubed over three. And we are going to evaluate that from zero to t. So that would give us a position function of negative, too. By filling t. I have nine t minus t cubed over three. Subtract what I get when I fill in zero. But when I fell in zero, I get zero so I can see that that is the exact same thing as I had before.

Bob Gorbold · Answer

So if we have a particle moving on a line like this, going laughed and right and all around, if we could describe its motion over to back one over to back one and we could define that as some function v of tea, then we would know, um, that if I want to know its displacement, I could simply take the integral of V of tea with respect to time. And that would give me the displacement between any two times. And so we just call him Time one time, 28 be something like that. So this Indyk role of velocity would give me the displacement because it would add up all the little net changes and give me an overall net change. And it will account for positives and negatives, just like we wanted to. And so if I knew the velocity of that red dot I could just integrated to get this placement

Robert Zaballa · Answer

given the following that a time equal to zero X. The position at time equals zero is equal to X, not in the velocity at time equals zero is equal to be not and that the acceleration is a function of time is given to a Z a knot plus b t where a not be are constants. We want to find the instantaneous velocity as a function of time and also the instantaneous position as a function of time. So we&#x27;re gonna use integration to find these functions. So we&#x27;re going to start by finding the velocity is a function of time, and the way we would do that is by using the integral of the acceleration. So we&#x27;ll start with this form and what the initial conditions. They&#x27;re going to be important in carrying out these in a girl&#x27;s, because the information that time equals zero is going to tell us what the constants of integration are. So we&#x27;re going to substitute in what we know and that is that acceleration as a function of time has the form specified in the given information. So right now, we simply have ah, a linear. We basically have a constant term plus in linear terms and in integrating a constant term, you obtain the constant times T you have a not times t and when you integrate the linear term, you have a quadratic term. So you have the integral B t is equal to B T squared over two. Now we have a constant of integration that comes out which for short, I simply right as constant like this. And the way we determine the value of that constant is we right via zero. And in doing so, this will take away these 1st 2 terms. T will be equal to zero and T Square will be equal to zero. So these 1st 2 terms will vanish and this will leave us with constant. But we already know that at time equals zero V 00 Levin Not so the constant of integration. Zika lovey Not and we then have for the instantaneous velocity is a function of time. The VT is equal to a not tea plus B t squared over two plus v. Not next. We&#x27;re gonna use this function, take its integral and obtain the instantaneous position as a function of time. So we have the integral of linear term. We have the integral of a quadratic term and we have the integral of a constant term. And we have DT This gives us a not t squared over two plus b t Cube and then we have two times three right here, which is six. And then we have the integral of the knot, which is simply v, not tea. And then we have again a constant of integration not to be confused with, So I&#x27;ve got to go back up. This constant of integration is not to be confused with the previous I just use the same notation because it means a constant of integration. Not not necessarily the same one is before. Let me make some room here and then we&#x27;ll just rewrite. So pause Veena T what&#x27;s that? Constant? Okay, so again, ex notch or basically X zero. In other words, a T zero x of zero equals X Not we know this because in the initial conditions above as we saw here, that is specified At T equals zero. When we look at what we just obtained for exit T the 1st 3 terms, they&#x27;re gonna vanish. Because if t is equal zero than a not t spread over two. Must be zero b t cubed over six Must be zero and v not team must be zero if t is equal zero. So that leaves us with the constant of integration. So our constant of integration for this function is equal to X. Not so or function is then fully determined. And we have a not d squared. Really. We can write this in order of descending powers of tea. So we will write this as BT cubed over six waas a not t squared over to most Veena t plus X not and we can do the same with our what we did obtain for the velocity. We can write these in descending order up here just to be consistent, we can say b t squared over two plus a not tea plus Veena and that&#x27;s what we have

Jonathon Brumley · Answer

we are going to calculate the position function two ways. First, we&#x27;re going to calculate it using the anti derivative method, and then we&#x27;re going to calculate it, using the fundamental theorem of calculus to find the position function using the anti derivative method. What we would do is simply integrate that velocity function. So we&#x27;re integrating sign T T. By definition, the integral of sine t d t is negative co sign t plus a constant. We do need to figure out what that constant ISS and what we were given should say. S zero equals one. We were given at the beginning that the position function at a time of zero was one. So if that&#x27;s true, then if I fill in zero for T, I should get one. So let&#x27;s do that. That would have that equals one coastline of zero is once I have negative one plus C equals one. If I add the one over see is too. So that tells me the position function Hiss Negative co sign T Plus two or to minus co sign t. One show that the fundamental thermal calculus yields the same thing. Fundamental thermal calculus would say that the position function It t is equal to the position. Function it zero plus the integral from zero to t of the velocity of X dx. Okay, well, we do know what the position Functional zero waas. It was defined to be one the velocity function at X which means office Fillon Expert e was given to be sine X. The integral of sine X is negative co Sainty so I&#x27;ll just put that negative out front. Now I have this and I&#x27;m a valuable dress Coastline acts not tea. Evaluating that from zero to t felon t that would give me co sign T minus co signed zero but coastline zeros is ah one. So I have one minus co sign T minus one. Distribute the negative. Oh, and what do I have? The exact same thing I had for part a tu minus co sign t

Averell Hause · Answer

So we have a function of our wife position. We have a function for the white position as a function of time and this is giving us 2.0 Sent two points or centimeters multiplied by sine of pi multiplied by t divided by four. And with respect to t, we can solved weaken differentiate to find the velocity in the UAE direction. With respect to t, this would be equal to the derivative of the why function with respect to time. And this is giving us pi over too centimeters per second and they&#x27;re gonna multiply this by co sign of pie t over for And so we consent Say that the average velocity rather you should get a new workbook for part of the average velocity. Is that gonna be equaling 21? And this would be for the 1st 2 seconds, so this would be one divided by 2.0 seconds. This would be multiplied by the Inter girl from 0 to 2 of the velocity in the white direction. With respect to time or D t, This is equal one over 2.0 seconds, multiplied by pi over two centimeters per second multiplied by the integral from 0 to 2 of co sign pie t over four d t. And so we can then say that the average velocity would be equaling two again, one over 2.0 seconds multiplied by two centimeters, multiplied by the integral from zero to pi over too co sign of Ex the ex, and this is equaling 1.0 centimeters per second. So this would be our answer for the average velocity for the 1st 2 seconds for a part for Part B, then the instantaneous velocities would simply be found by just a functional value. So you just plug in and we can say that the philosophy in the right direction AT T equaling zero seconds would be equaling pi over two centimeters per second, multiplied by co sign of zero and coast an observer. We know to be one, so this is equaling simply pi over two centimeters. For a second. The velocity in my direction AT T equal in 1.0 seconds would be equaling two again pi over two centimeters per second. This would be multiplied by co sign of pi over four coastline of power for is radical to over two and so we find that the velocity here would be pie radical to over four centimeters for a second and the velocity in the UAE direction AT T equaling 2.0 seconds. This is equaling again pi over two centimeters per second. However, here we have co sign of pi over to we know the coastline of pi over to is zero. Therefore, the velocity of the wind direction AT T equals 2.0 seconds would be zero centimeters per second. Four part See, we&#x27;re going to differentiate the velocity and direction with respect to time to find the acceleration With respect to time, this would be again, Devi said. Why? With respect to time, this would be equaling two negative pie squared over eight centimeters per second squared multiplied by sine of pie t over four. So to find the average acceleration. But for the 1st 2 seconds, we can say that continuing on her part. See, we have the average acceleration. This would be equaling 21 over 2.0 seconds. This would be multiplied by the integral from 0 to 2 of the acceleration in the UAE direction de ti equaling 1.2 point one over 2.0 seconds. Multiplied by negative pi squared over eight centimeters per second squared multiplied by the integral from 0 to 2 of sign pie T over four DT. And so the average acceleration is that gonna be equaling one over 2.0 seconds multiplied by negative pi over too centimeters per second. We&#x27;re going to multiply this by the integral from zero to pi over too. And this would be, of course, sign of X d x, and this is giving us one over 2.0 seconds multiplied by negative pie over too centimeters for a second. Given that this is one and so the average acceleration is then gonna be equaling negative pi over four centimeters per second squared. This would be the average acceleration for the 1st 2 seconds for part C four part D. Then we want the instantaneous accelerations at 01 and two seconds and we can say that here the acceleration in the UAE direction at T equal and zero seconds would be equaling. Two negative pi squared over eight centimeters per second, squared multiplied by sine of zero, which we know to be zero. And so this is equaling zero meters per second or other centimeters per second squared and the acceleration of the wire direction for tea equaling 1.0 seconds, this would be equaling. Two negative pi squared over eight centimeters per second, squared, multiplied by sine of pi over four. And so this is giving us negative tie squared, multiplied by radical, too, over 16 centimeters per second squared. And finally, the acceleration of the wire direction for tea equaling 2.0 seconds. This would be equaling two negative pi squared over eight centimeters per second, squared, multiplied by sine of pi over to which we know to be one. And so this is gonna give us negative pi squared over eight centimeters per second squared. That is the end of the solution. Thank you for watching

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

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Basic Technical Mathematics with Calculus