Velocity vector addition problem

AI Thread Summary
The discussion revolves around a velocity vector addition problem involving a boat crossing a river. The correct answer is a downstream angle of 53.13 degrees relative to the bank, derived from the vector sum of the boat's velocity and the river's current. Participants emphasize the importance of using vector addition correctly, noting that mixing reference frames leads to confusion. A specific geometry involving the river's width and distance along the bank is mentioned, with calculations indicating an angle of 26.565 degrees, which is not the final answer. The conversation highlights the necessity of understanding the problem's context and avoiding shortcuts in calculations.
evo13
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Homework Statement
Introduction physics
Relevant Equations
Vector addition
1.png


Hello, guys. Interesting riddle here.
I have no idea how to solve it. Tried different methods, but point is answer is always wrong,
exact answer Downriver, at an angle of 53.13(degree) to the bank.
That exercise is from
"Pohl’s Introduction to Physics"
 
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The basic idea is that the velocity of the boat relative to the banks is the vector sum of the velocity of the river relative to the banks and the velocity of the boat relative to the river.
 
Yes, and i can't get a right answer. Basic idea is that velocity of boat and river is the same magnitude, at least as i understand it
 
evo13 said:
Yes, and i can't get a right answer. Basic idea is that velocity of boat and river is the same magnitude, at least as i understand it
Let's see your calculations.
 
PS The given answer of 53.13 degress is correct.
 
1637226208606.png


Sorry for the sloppiness, this is just one of the solutions I tried
 
You know that the river is ##600m## wide and it's ##300m## along the bank from ##A## to ##B##. The angle ##\alpha## can be calculated from this geometry.

I.e. ##\tan \alpha = 1/2##

You should get the answer from that.
 
PeroK said:
You know that the river is ##600m## wide and it's ##300m## along the bank from ##A## to ##B##. The angle ##\alpha## can be calculated from this geometry.

I.e. ##\tan \alpha = 1/2##

You should get the answer from that.

Yep, i did that. ∠α is 26,565 degrees
AB = sqrt(45) is not relevant, sorry.
 
evo13 said:
Yep, i did that. ∠α is 26,565 degrees
AB = sqrt(45) is not relevant, sorry.
That gives you the direction relative to the bank. The question wants the direction relative to the water, which means you have to do vector addition.
 
  • #10
Sorry but vector addition is second part of my solving attempt, below horizontal line.
I don't understand what else i can do
 
  • #11
PS I don't see how you got 58.3 degrees. You should have got 63.4 degrees upstream as the direction relative to the bank. But, that is independent of the velocity of the river.
 
  • #12
lets try another explanation, more clear i hope
c5cbf771-1fba-492b-bee5-20fdedb30cc2.jpg
 
  • #13
Okay, I see what you've done. Why would ##a = b##?

Note that your diagram is based on the reference frame of the banks. It's difficult to indicate the velocity of the boat relative to the river on that diagram.

Vector addition is the way to go!
 
  • #14
if not a = b, than c = b. Right?
I tried this also
 
  • #15
evo13 said:
if not a = b, than c = b. Right?
I tried this also
Neither. Your fundamental problem is that ##c## is from the river's frame of reference and ##a## is from the banks frame of reference. You are mixing vectors from two reference frames in one diagram.

Even if these equations were to hold, they can't be shown in a single diagram.

Use vector addition!
 
  • #16
Ye, alright. Thank for the help.
I am clearly not understand the question than, English is not my native.
 
  • #17
evo13 said:
Ye, alright. Thank for the help.
I am clearly not understand the question than, English is not my native.
You do understand the question. You're trying a clever shorcut that doesn't work!
 
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