Solving a River Boat Problem: Distance Downstream

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Homework Help Overview

The problem involves a boat crossing a river while being affected by the current. The river flows due North at 3.17 m/s, and the boat travels due East at 8.5 m/s relative to the water. The objective is to determine how far downstream the boat is when it reaches the opposite shore, which is 163 m wide.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the resultant velocity and its direction. There is a mention of using the law of motion, with some questioning whether the approach is appropriate for uniform motion. Others suggest using geometry instead of finding time.

Discussion Status

The discussion includes various attempts to apply different methods to solve the problem. Some participants express confusion about the appropriateness of using accelerated motion equations, while others propose a geometric approach to find the downstream distance.

Contextual Notes

There is a mention of uniform motion and the need to clarify the assumptions regarding the motion of the boat and the river current. The participants are navigating through different interpretations of the problem setup.

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Homework Statement


A river flows due North at 3.17 m/s. A boat
crosses the river from the West shore to the
East shore by maintaining a constant velocity
of 8.5 m/s due East relative to the water.
If the river is 163 m wide, how far downstream
is the boat when it reaches the East shore?
Answer in units of m




The attempt at a solution
I found the magnitude of velocity which was 9.7187 and its direction.
I used the law of motion
y)163=3.17sin(90)t+4.9t^2
to solve for x) x=8.5cos(0)(time i got for y)+4.9(time i got for y^2)
Am I going about this the right way?
 
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Hi blayman5! :smile:
blayman5 said:
I found the magnitude of velocity which was 9.7187 and its direction.

hmm … I make it 9.07187.
I used the law of motion
y)163=3.17sin(90)t+4.9t^2
to solve for x) x=8.5cos(0)(time i got for y)+4.9(time i got for y^2)
Am I going about this the right way?

Nooo … that looks like an accelerated motion equation. :confused:

i] This is uniform motion

ii] You don't need to find the time …

you know the direction from the first part … so just use geometry! :smile:
 
So use the magnitude of velocity and direction along with the water velocity and solve for the x side?
 
i get it

tanx=length/width
width*tanX=length

163*tan(20.4526)=60.7896
thanks
 

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