Magnitude and Direction, Canoe on River

AI Thread Summary
The discussion revolves around calculating the velocity of a canoe relative to a river, given its velocity relative to the earth and the river's current. The canoe's velocity is 0.35 m/s southwest, while the river flows at 0.65 m/s east. The attempted solution indicates a magnitude of approximately 0.93 m/s, but the angle relative to east remains unclear. Participants emphasize the importance of understanding vector addition and the use of x and y components to determine angles. The conversation highlights the need for clarity in vector calculations to solve the problem accurately.
ap_physics
Messages
2
Reaction score
0

Homework Statement


A canoe has a velocity of 0.35 m/s southwest relative to the earth. The canoe is on a river that is flowing 0.65 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

The Attempt at a Solution


0.930985 is the magnitude.
Then I have no clue how to get the angle in terms of counterclockwise from east.

Thanks very much in advance!
 
Last edited:
Physics news on Phys.org
ap_physics said:

Homework Statement


A canoe has a velocity of 0.35 m/s southwest relative to the earth. The canoe is on a river that is flowing 0.65 m/s east relative to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

The Attempt at a Solution


0.930985 is the magnitude.
Then I have no clue how to get the angle in terms of counterclockwise from east.

Thanks very much in advance!

Hi ap_physics; Welcome to Physics Forums.

How did you arrive at an angle of 315 degrees for the canoe's velocity? I would have thought that "southwest" would be halfway between south and west.

Edit: I see you've edited your original so that the angles of the vectors are no longer shown. Perhaps you could just explain your approach to the problem?
 
Well, I posted my attempt earlier, but it was wrong, and I was given the answer for the velocity by my teacher, as it was too complicated for our level. But, he says I can still find the angle. But, I don't know which formula to use.
 
ap_physics said:
Well, I posted my attempt earlier, but it was wrong, and I was given the answer for the velocity by my teacher, as it was too complicated for our level. But, he says I can still find the angle. But, I don't know which formula to use.

Okay, what have you learned so far about adding and subtracting vectors? Have you learned how to determine their angles from their x and y components?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top