Question

A canoe has a velocity of 0.390 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.760 m/s east relative to the earth.

Find the magnitude of the velocity v⃗ c/r of the canoe relative to the river

Answer 1

Given Velocity of canoe relative to earth = Vce = 0.390 m/sec. Southeast relative to the earth

Vce = 0.390 m/sec. at 45 deg south of east

Let x-direction will be in east direction and positive y-direction be in north direction.

So, Vce = 0.390*cos(45 deg) i + 0.390*sin(45 deg) (-j)

Vce = 0.276 i - 0.276 j

also, velocity of river = Vre = 0.760 m/s in east direction

Vre = 0.760 i

Now using relative velocity,

Vce = Vcr + Vre

here, Vcr = velocity of canoe relative to river = ??

Vcr = Vce - Vre

Vcr = (0.276 i - 0.276 j) - 0.760 i

Vcr = (0.276 - 0.760) i - 0.276 j

Vcr = -0.484 i - 0.276 j

magnitude = |Vcr| = sqrt(0.484^2 + 0.276^2)

|Vcr| = 0.557 m/sec.

direction = = arctan(Vy/Vx)

= arctan(0.276/0.484)

= 29.69 deg

therefore velocity Vcr of the canoe relative to the river at 30 deg at south of west.

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