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This is a symbolic version of Problem 29. A river has a steady speed of $v_{s}$ A student swims upstream a distance $d$ and back to the starting point. (a) If the student can swim at a speed of $v$ in still water, how much time $t_{\text { up }}$ does it take the student to swim upstream a distance $d ?$ Express the answer in terms of $d, v,$ and $v_{s .}$ (b) Using the same variables, how much time $t_{\text { down }}$ does it take to swim back downstream to the starting point? (c) Sum the answers found in parts (a) and (b) and show that the time $t_{a}$ required for the whole trip can be written as

$$t_{a}=\frac{2 d / v}{1-v_{s}^{2} / v^{2}}$$

(d) How much time $t_{b}$ does the trip take in still water?

(e) Which is larger, $t_{a}$ or $t_{b}^{2}$ Is it always larger?

a. $\frac { d } { v - v _ { s } }$

b. $\frac { d } { v + v _ { s } }$

c. click for proof

d. $\frac { 2 d } { v }$

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Numerade Educator

University of Washington

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Hope College

Hello, everyone and this problem. Whereas, to find the time it takes for a student to swim upstream off reverb that is flowing with a constant speed B s. The student has a maximal speed off the in the water and so first were asked to find What if the student is swimming up stream? Uh, this is the okay. And so this trip, assuming that the students starts from rest, takes a distance, that is, uh, or takes a time. That is, the net equals deal over t up and so t up is the over the net, which is V minus B stream now here have chosen upstream. So the direction that the student wishes to swim as the positive direction. And that means that in this direction, the net velocity or the speed that the student is going to have is is his or his or her own speed minus the speed of the river. And so that's how I got the net over here. And so t up is deep over V minus b s. For the second case, the student is swimming downstream. So in this case, we choose downstream as the positive direction and then Vienna Prime is going to be equal to the sum of the two speeds again, which is the students speed itself in Stillwater, plus thes streams speed. And this is going to be equal to the same distance covered over the new time taken for for the student to cover this distance which is t down or TV and so t d is equal to d over the net prime which is equal to 3/3 plus ves. So notice that now the speeds at up for the student, right? So the river is helping the student swim or get to the point where they started from faster. So then adding the two we find that t a which is one part of the or one way to do the trip which is in a flowing river. So this is Thea plus D d is D overview managed DS plus D over B plus bs. Now we bring everything that you common denominator which is one over re squared, minus to be a squid. So this is just multiplying this first term by the plus ves over V plus V s and the second term by the minus V s over B minus V s and in fact, arising the over three squared minus B squared. And so we're left with in the brackets with we Plus we s plus V minus V s where noticed that the streams speeds canceled its you're left with two V D over V squared, minus b squared. But then fact arising the square from that you get to VD over B squared times one minus B squared over b squared. And here, um, this via in the numerator cancels one of the factors of the V in the denominator. And so eventually you're left with two times dealer V overall minus via squared over B squared. So this is the final wrestle for for completing this trip. Then pretty, you're asked what happens, Or how long does the trip take it? Still water. So that case B s zero so immediately You could just set V s equal to zero here and you would find the T B is equal to to the over V or alternatively, you could say that the speed of the student is can be measured as two times the distance. So the student goes up and down so covers twice the distance of the in a time TV. And so you could express TB that way, with the same result to the Overbey. And so part is asking which one is bigger? So now you're comparing to the over V over one minus B squared over a squared B s squared over B squared on to the over V. Now what happens? Um, you asked which one is larger, right. So let's think about this, um, in three different cases in terms off the relative velocity between or the relative speed between the river and the students. So if the student has a speech that is equal to the streams speed, then ves over V squared becomes one or approaches one. In that case, the denominator approaches zero, Which means that t a sort of trip to complete this trip, the sort of time taken to complete the trip up and down the river approaches infinity. Now, this makes sense because if you're on lee swimming or you can Onley swim as fast as the river is flowing, then you're essentially staying in place. You're not moving relative the river. You cannot essentially cannot swim up the river on the same thing happens if you have a speed that is less than the streams speed. In this case, you cannot. You can't just not swim off upstream. You're actually, uh, sway or are being are floating downstream as you try to swim across or you try to swim against the water. So these air to cases, in which case you cannot, you literally cannot do. You cannot even make the trip that we talked about in part A and beach. Um, now, what happens if the speed with which you swim route that the water is greater than the speed of the water? Well, in that case, what happens is that a V over V s is gonna be smaller than one. So this guy so this is smaller than one. If this is smaller than one, then this fraction is also smaller than one. So if that's smaller than one, then you're dividing by less. So you get a bigger a bigger time, right? So one over something smaller than one. Is this one over? Let's say you know this is 0.7 or something. Now, this is gonna be this is gonna be greater than one. So if this happens, then you're going to get that t A. So the time it takes to complete the trip when the river is flowing is longer than the time it takes to complete the trip when the river is not flowing.