Swimmer in river relative motion crap

A river 570 ft wide flows with a speed of 6 ft/s with respect to the earth. A woman swims with a speed of 3 ft/s with respect to the water.

a) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?

b) If she wants to be swept a smaller distance downstream, she heads a bit upstream. If she heads 35° upstream, how far downstream is she swept before reaching the opposite bank?

c) For the conditions of part (b), how long does it take for her to reach the opposite bank?

I'm stuck on part b only. :-(

Here are some equations I figured out from my book.

xf= t * vx = ( d / vy )* vx
vy = vA*cos (theta)
vx = vA*sin(theta) + vC

where vA is the velocity of the swimmer
Vc is the velocity of the river
vy is the y component of velocity
vx is the x component of velocity
xf is final position

So for part a) I did position=(d/vy)*(vx) and the angle is zero, so
(570ft/3ft)(6ft)= 1140 ft, which is right.

Part c)
I did:
vy=vA*cos theta
vy= 3*cos 35 =2.457 m/s
then t= d/vy => (570ft)/(2.457 m/s)= 231 seconds.

Now for part b:
I tried this:
vx=vA*sin theta+vC
vx=3*sin 35 + 6 =>7.720
vy=3*cos 35 =>2.457
position=(d/vy)*(vx)=> (570/2.457)(7.720)=1790.96 ft

Thing is, computer won't take my answer, and I believe my answer is correct. Hopefully there isn't a bug in the computer program. Is there another way of doing this problem, maybe easier?
Last edited:
For part b, vx due to the river's velocity is 6 ft/s, and vx due to the swimmer's x-component is -3sin35 (she's heading against the current in the x-direction, right?).

So what's the net velocity in the x-direction? Multiply that by the time it takes to get across.

Also, check your rounding of the time calculation.
Big thanks! 1 more thing

That was awesome. I figured out part b, thanks to you guys. Anyway, for part a) is there, perhaps, another way, maybe easier way of approaching the problem? I don't want to be stuck memorizing equations and stuff for the test.
I don't really know what you mean about memorizing formulas, especially in part A. Just think about what's happening.

One thing that you can't get away from in these problems is that you have to resolve the velocities into their x- and y- components. Then remember that the total velocity in the x direction is simply the sum of the x-components, and similarly for the y direction.

But in part A, there really are no components. The swimmer's velocity is ONLY in the y direction, and the river's velocity is ONLY in the x direction. So what do you know? You know the distance to be covered in the y direction, and you know the velocity in the y direction, so you just divide one by the other to find out the time it takes to get across. Then, since you also know the river's velocity in the x direction, you multiply that velocity by the swimming time to find out how much distance is covered in the x direction in that amount of time. Why do you need a formula for that?

If you're talking about using sines and cosines to get the components in part B, you pretty much can't get away from that. But don't do it by memorizing formulas. Learn to draw yourself a diagram showing the direction of the known velocity, and the direction(s) of the component(s) that you need to solve the problem. Build a right triangle around those elements, identify the angle between the components you need, and then teach yourself how to use the appropriate trig functions to get the values you need. If you hope to be successful in your physics course, it is CRUCIAL that you learn RIGHT NOW to do this by reasoning, and not by memorizing formulas. You will need to apply this same approach to all kinds of problems throughout the course, and if you try to do it by just memorizing the formulas to use for dozens of different situations, you'll suffer. Invest whatever amount of time it takes right now to get COMPLETELY comfortable with how to reason your way through these problems using basic trigonometry rather than formulas, and it will make your life a lot easier throughout this semester (and the next).

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