Calculating Relative Velocity of a Canoe on a River

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The discussion focuses on calculating the velocity of a canoe relative to a river, given its velocity relative to the earth and the river's flow. The canoe moves at 0.30 m/s northwest, while the river flows at 0.50 m/s west. The solution involves using the equation V_{C/E} = V_{C/R} + V_{R/E} and applying the Pythagorean theorem to find the resultant velocity. The final answer is determined to be 0.36 m/s at an angle of 53.6 degrees east of north. The calculations clarify the components of the velocities involved in the problem.
Edwardo_Elric
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Homework Statement


A canoe has a velocity of 0.30m/s northwest relative to the earth. The canoe is on a river that is flowing 0.50m/s west relative to the earth. Find the velocity (magnitude and direction ) of the canoe relative to the river.




Homework Equations


E = earth
R = river
C = canoe
V_{C/E} = V_{C/R} + V_{R/E}
V_{C/E} - V_{R/E} = V_{C/R}

The Attempt at a Solution


the answer at the back of the book is .36m/s 53.6 degrees east of north

to get to the book's answer, i used the pythagorean theorem
\sqrt{{V_{C/E}}^2 - {V_{R/E}}^2} = V_{C/R}
and i confusingly substituted
\sqrt{((0.30)cos(45) - .50)^2 - ((.30) sin 45))^2} = V_{C/R}
then get .36
and i don't know how the V_{C/E}'s x components got involved ...
 
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ohhh
ok i got it already:
since : V_{C/E} - V_{R/E} = V_{C/R}
x component: -(.21)- (-.50)
y component: .21 - 0
(.29)^2 + (.21)^2 = V_resultant^2
0.36

\theta = arctan\frac{.21}{.29}
\theta = 35.9degres north of east
or 90 - 35.9 = 54.1 e of n
 
I like these kind of threads! :biggrin:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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