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(20 pts) A river flows northeast at 1 m/s. A canoe is swept along with the current with an apparent velocity of 0.2 m/s west.
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Answer #1

here,

the speed of river relative to ground , vrg = 1 m/s * ( cos(45) i + sin(45) j)

vrg = 0.71 i m/s + 0.71 j m/s

the speed of canoe relative to ground , vcg = - 0.2 m/s i

the velocity of canoe relative to river , v = vcg - vrg

v = (-0.2 i) - (0.71 i + 0.71 j) m/s

v = ( - 0.91 i m/s - 0.71 j m/s)

the magnitude of velocity of canoe relative to river , |v| = sqrt(0.91^2 + 0.71^2) = 1.15 m/s

the direction of velocity of canoe relative to river , theta = arctan(0.71/0.91) south of West

theta = 37.96 degree South of West

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