Question

A canoe has a velocity of 0.31m/s southeast relative to the earth. The canoe is on a river that is flowing 0.48m/s east relative to the earth.

Find the magnitude of the velocity of the canoe relative to the river.

Find the direction of the velocity of the canoe relative to the river.

Answer 1

A canoe has a velocity of 0.310 m/s southeast relative to the earth
@ 315 degrees CCW x-axis: vcg

A river that is flowing at 0.480 m/s east relative to the earth
@ 0 degrees CCW, but ***-vrg now @ 180 degrees***

Relative velocity "vcr" in this case is the vector subtraction of the velocities of the objects:vcg-vrg

The velocity of the canoe relative to the river.
where g=ground
vcr=vcg+(-vrg)

vx
0 .31* Cos(315)+ 0.48* Cos(180) =-0.26i m/s

vy
0 .31* Sin(315)+ 0.48* Sin(180)=-0.21 j m/s

1. Find the magnitude of the velocity of the canoe relative to the river.
vcr=sqr(vx^2+vy^2)=magnitude = 0.334 m/s

Angle positive x-axis:=atan(vy/vx) =38.9 degrees

Answer 2

Vc = 0.31*cos45i - 0.31*sin45 j

vcE = 0.22i - 0.22j

vrE = 0.48 i

VCr = VcE - VrE = 0.22i - 0.22j-0.48i

Vcr = -0.26i - 0.22j

l Vcr l = sqrt(0.26^2+.22^2) = 0.34 m/s

direction

tan^-1(0.22/0.26) = 40.23 with south of west

Answer 3

Vc = 0.31cos(45

Answer 4

Part A = 0.1733 m/s
Part B = 27.67