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A river flows due south with a speed of 2.30 m/s. A man steers a motorboat across the river; his velocity relative to the water is 4.50 m/s due east. The river is 900 m wide.

In which direction should the motorboat head in order to reach a point on the opposite bank directly east from the starting point? (The boat's speed relative to the water remains 4.50 m/s.)

What is the velocity of the boat relative to the earth?

How much time is required to cross the river?

I thought someone provided a pretty readable explanation on this below. I have the time to work on this too, but I'd like to know where the difficulties lie.

Let's use the standard x-y coordinate system to analyze the problem.
The river's velocity is 2.30m/s due south so it's velocity vector looks like

The boat's velocity is 4.50m/s due east so the boat's velocity vector looks like

The distance across the river is 900m. Hopefully you can see that if the man were to just aim the boat due east he would arrive downstream from where he started. If we were to add the velocity vectors together we'd get the direction he'd be traveling as
boat's vector = (0,-2.3m/s) + (4.5m/s,0) = (4.5m/s,-2.3m/s)
For each 4.5m he goes east, he'd go 2.3m south. Can you draw a diagram or picture what is taking place here?
What angle should he use to compensate for the river's velocity? What we want is the vector he should use so he ends at the point (900,0) (Keep in mind that position vectors and velocity vectors are two different concepts, but we can plot them in the same system.)
If were to add the vector that is the negative of the river's velocity to the due east vector, we'd have
(4.5m/s,0) + (0,2.3m/s) = (4.5m/s, 2.3m/s)
If we now add the river's velocity to this, we get the boat's vector as
boat's vector= (4.5m/s, 2.3m/s) + (0,-2.3m/s) = (4.5m/s, 0)
This means that for every 4.5m he goes east, he should go 2.3m north to compensate for the river flow. This angle would be
tan(theta)=2.3/4.5 so theta=arctan(23/45) = approx 27.1deg from due east.
The component of the boat's velocity that is moving due east is
cos(27.1deg)*4.5m/s=approx 4.01m/s
How long does it take to cross? 900m=cos(27.1deg)*4.5m/s * t = 4.01m/s * t
Solve for t.
As for the velocity relative to the earh, look at where it starts and where it ends. What do you think? Let me put it this way. If the boat needed to go upstream, this vector would make no progress doing that.

I think I solved this problem incorrectly for the angle and that affects everything.
We know the angle theta is north of east so the boat's velocity vector is
boat vector = (cos(theta)*4.5m/s, sin(theta)*4.5m/s)
When this is added to the river's velocity we want the y components to cancel, thus we want
sin(theta)*4.5m/s=2.3m/s so sin(theta)=2.3/4.5 =approx 30.7deg
Thus the correct velocity vector for the boat is
Boat_velocity=(cos(30.7)*4.5m/s, sin(30.7)*4.5m/s)

However, relative to the earth the boat is still going due east, but it's relative velocity is 3.87m/s due east.
The time needed is
900m=cos(30.7deg)*4.5m/s * t = 3.87m/s * t, now solve for t.
I realized after I logged off that I hadn't determined a unit vector in the boat's direction. When I did this I realized I interpreted the problem incorrectly. I hope whoever asked this reviews this post. Ordinarily I review my post about an hour after I post them for accuracy, and if I find an error I try to post the correction when I can. I certainly appreciate others reviewing and correcting incorrect responses too.

Thanks Roger. I actually figured out that you solved it wrong, and did the work based on your formulas. Got the answers right. Thanks.

I'm glad to hear that. I just hope it didn't cause too much confusion in learning the right method.

Thanks Roger. I actually figured out that you solved it wrong, and did the work based on your formulas. Got the answers right. Thanks.

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