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For $t<0,$ an object of mass $m$ experiences no force and moves in the positive $x$ direction with a constant speed $v_{i}$ . Beginning at $t=0,$ when the object passes position $x=0$ , it experiences a net resistive force proportional to the square of its speed: $\overrightarrow{\mathbf{F}}_{\text { net }}= m k v^{2} \hat{\mathbf{i}},$ where $k$ is a constant. The speed of the object after $t=0$ is given by $v=v_{i} /\left(1+k v_{i} t\right)$

(a) Find the position $x$ of the object as a function of time.

(b) Find the object's velocity as a function of position.

$V_{1} e^{-k x}=V$

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Okay, So this question we have an object and it has a net resisted for us proportional to the square of the speed. So let's focus on the net. Resistive force. Um, I think you should be negative, but I'm gonna write like this for now. Maybe leave a little space for a negative. So is it m k B squared in the eye? Hot direction. So, um, the speed is v initial divided by one plus k b t throwing at the position and velocity as a function of position. Ok, well, first, we want to get position as a function of time. So write that down. X t equals question. Mark. I'm a little troubled by the sign because we know that if v is positive, um uh, actually, you positive think. Well, if we know the velocity of the function of time, we should also be able to get position by observing that velocity is DX DT. So then, um, X is a function of t is going to be, um uh do you kind of integrate both bring the DT over? So I'll cost Cross that out, and then you want to integrate both sides and um, And so the left this side becomes just acts like the position you can say starts at ax equals zero to some X equals X final. Um and I guess to be more clean, I can use an integration variable dx prime, whatever. But this whole this side, we get X and then this side when I integrate with respect to time. So on the integral of so we can kind of do a U substitution thing. So you is k the initial time. Do you d t is equal to cave be initial. Oh, plus one. Um, so during that you substitution means I'm doing the integral of one over you, which is just going to be the l. And, um um, you and I need to suffer DT kind of doing this whole once, but I'm assuming you've seen in a girl's before. You complete this interval from Alfa as well. Um, believe me, or maybe just to check me, I haven't done in a girl's much recently either. Okay. And then we wanted someone for DT is, um do you over Katie, I Okay, so that's X is a function of time. Is there any simplification I guess I can cancel out this initial velocity. So I have the Alan of Cavey initial time plus one divided by the initial who works. OK, I guess K has to have inverse units of Yeah, okay. Has units of inverse meters because of velocity. Times time, Miss Meters. So K's inverse meter. So it actually works out that it's one over k. So that's X. But everyone X is a function of V. So we need to eliminate time. So, um, so we need to take V in sulphur time, so I'll rewrite be over here. So one plus k the initial t. But I want to stall for time something, um, multiply the top and bottom by one plus k v initial t So I get V times one plus que the initial t equals B. I and, um, yeah, I'm just gonna be drink algebra from here on out. So, um so divide by B and subtract one. So be initial over. B minus one is equal to K B initial t, and then you get that t is equal. 21 over k. Okay, be initial, um, times v I over B minus one and then I want to take that, unplug it and hear someone over K Time C l and, uh, Lewis is ugly. Um, the cavey initials are gonna cancel when I said the t so Oh, this isn't so bad. And I have the initial over B minus one. Um, nice. Um oh, and then plus one. So then that so then that's going to just give me, um, the initial Overbey because this minus one here canceled with this plus one here. I'm right out yourself a good if it does not make sense. Um, and I was kind of hard to listen to somebody else's algebra, so yeah, that's position as a function of velocity. Then only one velocity is a function position. Okay, fine. So let's actually solve Ruby. So then we'll have, um so bring this k over. So okay, X is the Ellen of this stuff, and then we exponentially eight both sides and then we get it's equal to be I Overbey. So we get V is equal to be i e to the minus k x. Um, just by bringing the be over and then dividing right of the K axe and then dividing, right, you the kiosks means he becomes e to the negative K ox. The newer