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Numerade Educator

# (II) An object moving vertically has $\vec{\mathbf{v}}=\vec{\mathbf{v}}_{0}$ at $t=0$ .Determine a formula for its velocity as a function of timeassuming a resistive force $F=-b v$ as well as gravity fortwo cases: $(a) \vec{\mathbf{v}}_{0}$ is downward and $(b) \vec{\mathbf{v}}_{0}$ is upward.

## a. $v=\frac{m g}{b}\left(1-e^{-\frac{b}{m} t}\right)+v_{0} e^{-\frac{b}{m} t}$b. $\frac{m g}{b}\left(e^{-\frac{b}{m} t}-1\right)+v_{0} e^{-\frac{b}{m} t}$

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##### Farnaz M.

Simon Fraser University

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg

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### Video Transcript

in this question, we have an object moving vertically with a velocity of V zero at time equals zero and were asked to determine a formula for the velocity as a function of time. Assuming that there is a resistive drag force as well as gravity acting on the object in case A we're going to assume that the initial velocity is downward and in case B will assume the initial velocity is upward. So I think the first difficult thing about this particular question is we want to find this expression for velocity, but it's not exactly clear where to start. So, you know, one might be drawn to trying to use Akina Matics approach at first because we have a lot of expressions for V in arcana Matics toolbox. The problem with that is that those equations are developed under the, um, assumption that acceleration is a constant, and in this case, that's just not the case. Um, so it turns out that the best possible method here, um, is to approach from a dynamics aspect. So we're going to start off by considering a downward velocity initially, and I'm actually just to make things a little bit easier. I'm just going to assume that this is the downward direction is the positive direction. And we know that if this objects yeah, is flying through the air, it's going to have a downward gravitational force, equal toe MGI and an upwards drag force equal Thio, BB. And the reason why the drag forces upwards is because this object is moving downwards. And so the drag force is always going to try to act, too. Um, resist that motion. So it's going to be upwards in this case. So as I said, we're going to take a dynamics approach to this and we're going to create an F not equals. I'm a equation. Which efforts is going to seem a little bit strange, but it becomes pretty clear why we're doing this quickly. So the forces on this object are BV positive. Um, sorry, Negative BV, because we're taking up to be the negative direction. Plus positive MGI equals m a now because I want to have an equation for velocity. I'm actually going to replace a with D v d t. And that gives me a differential equation for V, which we can solve, um, in order to solve this equation, we're going to be using a method called the integrating Factor. Unfortunately, we can't use the separation of variables method here because this equation is not separable. So if you don't know the integrating factor method or you haven't used it in a while, you might wanna do a little brush up. Now, in order to use integrating factor method, we're gonna have to rearrange this equation a little bit. So I'm going to take my M D v d t to the other side. I'm going to keep the minus BV there, and I'm going to take the mind the mg over to the other side and then to use integrating factor DVD T should be by itself. So we're going to divide everything by negative M. So that's going to give DVD t plus B over em the, um And if we're dividing by ah, negative M, then we will just get she. So the reason that I rewritten it like this is because to use integrating factor, the D v d T terms should be by itself. Um And then we could be on the same side of the equation. Is that and then any Constance will go to the right hand side of the equation. So the next step is to calculate the actual integrating factor. And that's gonna be e to the power of an integral. The integral we're going to take is of whatever is in front of the V. So this integral is going to be the integral of be over m d t. So that's just an integral of a constant. So we get e to the power of be over em t once we complete diet integral and then in the integrating factor method, What you're going to dio is you're going to multiply everything by the integrating factor. So every term is gonna have this exponential in front of it. And once we've done that, we're going to go ahead and integrate both sides. This is where the magic happens, because you'll notice that Oops, I forgot to be over. I'm here. You'll notice that on the left hand side here, what we have looks kind of like a product rule. Um, so this if you think backwards is actually just the integral of the integrating factor e to the power bi over MT. Times V and So when we take the derivative Oh, sorry. When we take the integral of the derivative here, that all goes away and on the right hand side or left hand side, we're just left with e to the power bi over m t times V. And that's gonna be equal to that integral. So that kind of really cleans up the left hand side, that little trick quite a bit. And that's what integrating factor method is all about. So I'm just going to go ahead and take this integral should be eternally be a DT here. And, um, the integral of e to the B over m t is basically just itself. But we do need Thio take into account the fact that if you were to differentiate this and try to get back to the original, um, a B over em would come down. And so I need to place an M over be out front to cancel that out, Okay. And of course, um, since this is indefinite integral, we need to add on, uh, plus C for the constant. And finally, since we're trying to create an equation for V, I'm just going to divide by the integrating factor. So we've got we'll end up with MGI, Overbey plus C E to the minus Be over MT. And in order to get rid of this see, we can substitute in our initial conditions. So we know that V is equal to V not at time equals zero. So I'm going to sub in V not for V and zero for team. When I do that, the e to the minus B over m t is going to become one. Eat the power zero is just one. So I get this and we can rearrange for see, So that's going to be V zero mg Overbey. So I'm gonna plug the constant back into this equation here, and we'll get our final expression. Now we can play around with this a little bit to make it look a bit nicer. So I'm gonna put the I'm gonna expand out the bracket and I'm gonna put the MGs together. So that's going to be MGB plus MGB. Eat the minus B over em T um plus V zero e to the minus B over mt. And then just factor out an MGB from the first two terms. Uh, Oops. So that will be one minus or sorry, one. This should be a minus. Here we go. One minus E to the minus. B over empty plus V zero each, the minus B over. Empty. So this is our final expression here for the velocity when the initial velocity is downward. And then we will do the upward case in a very similar manner. So this is for a part B, and we're going to assume that V zero is upward this time. I'm going to take up to be positive, because it's just easier to assume that the upward direction is the same direction as the initial velocity. Um, so that means that this time the forces are both going to be negative. Both the drag force and the, um, gravitational force are both gonna be pointed down. Um, this is because the object is moving up, so the drag force must be pointing down to resist that motion. And we're going to create a very similar equation. F not is equal to m A, and we can stop everything in making the same substitution as the previous part. And again, we want Thio solve this differential equation. I'm we're going to use integrating factor method. Um, so I'm gonna rearrange in a very similar way here. I'm going to divide by negative. Mm. This time when I do that, I'm going to get and negative g on the right hand side. So it's basically the exact same equation. Um, the integrating factor is the same. So it's, um, each of the power bi over Mt. And I multiply everything by the integrating factor. Um, so oh, this should be over. Beat over. I'm sorry. So once we integrate both sides, um, we're going to get very similar to last time e to the power bi over MT. Times V. And then on the right hand side, we're just going to get this integral again. So then we'll get e to the power. Bi M T. V is equal to negative mg Overbey. And, um, don't forget the plus c. Okay, so since we want an expression for V again, we're gonna just divide everything by e to the be over Mt um so that we get negative mg Overbey plus see each of the minus B over empty and very similarly to the last time we go ahead and we sub in our initial condition. So that will give us V zero is equal Thio minus mg over B plus C. And so C is equal Thio v zero plus m g over b and we go ahead and set that back into the full equation. So that's going to give us a V zero is equal to negative GM over B plus V zero mg Overbey that's all multiplied by the exponential term. And then just to make this look a little bit nicer, we can go ahead and, um, we can go ahead and multiply in the exponential and just do a little bit of rearranging. So once we do that, we're going to get em. G Overbey one minus e Oh, sorry. E to the minus. B over MT minus one for this one plus v zero times exponential term. So similar, but slightly different to part A. And so this is the final answer for apart beat

McMaster University
LB
##### Farnaz M.

Simon Fraser University

##### Aspen F.

University of Sheffield

##### Jared E.

University of Winnipeg

Lectures

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