A canoe has a velocity of 0.480m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.510m/s east relative to the earth. Find the magnitude of the velocity (V(C/R)) of the canoe relative to the river. Also, find the direction of the velocity of the canoe relative to the river.
The picture shows V(R/E) pointing east along the x-axis and V(C/E)pointing southeast, 45 degrees below the x-axis.
What I've tried to do:
V(C/R) = ?
V(C/E) = 0.48 m/s SE
V(R/E) = 0.51 m/s E
V(C/R) = V(C/E) - V(R/E)
Then I split each into their x and y components and added the x's together and y's together and took the square root of the sum of the squares:
V(C/E)---------------x: 0.48 cos 45----------------y: 0.48 sin 45
V(R/E)---------------x: 0.51 cos 0 -------------------y: 0.51 sin 0
Added together---x's: 0.849411-------------------y's: -0.339411
Sqrt(0.849411^2 + 0.339411^2) = 0.914712
So shouldn't this be my answer? What am I doing wrong?
Help please!
This is online homework and here are my answers that I've tried so far along with any feedback that they have given me:
0.172 m/s WRONG
0.361 m/s WRONG A vector that points southeast is inclined 45 degrees south of east
0.915 m/s WRONG Recall that the velocity of the canoe relative to the earth is equal to the sum of its velocity relative to the river and the velocity of the river current relative to the earth.
0.914 m/s WRONG
1 answer
2 answers