A canoe has a velocity of 0.460 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.600 m/s east relative to the earth. Find the magnitude of the velocity of the canoe relative to the river.

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Set up a vector diagram to calculate the velocity of the canoe relative to the river. 

To do this use the relative motion formula.

`V_(ac)=V_(ab)+V_(bc)`

Put into words, the velocity of a with respect to c is equal to the velocity of a with respect to b plus the velocity...

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Set up a vector diagram to calculate the velocity of the canoe relative to the river. 

To do this use the relative motion formula.

`V_(ac)=V_(ab)+V_(bc)`

Put into words, the velocity of a with respect to c is equal to the velocity of a with respect to b plus the velocity of b with respect to b.

You will find:

`V_(canoe,river)+V_(river,earth)=V_(canoe,earth)`

Where we dont know `V_(canoe,river)`

We have 2 lengths of the triangle, and the angle between those 2 lengths. For simplicity, lets now call the lengths we do know b and c and the angle between them A. 

The length opposite the angle A is what we require as this is the magnitude of the velocity of the canoe (namely, its speed) relative to the river.

We can calculate the length a using the Cosine Rule:

a^2 = b^2 + c^2 - (2bc cosA)

Giving

a = sqrt (b^2 + c^2 - 2bc cos A)

Now, we have that the lengths b and c are 

b = 0.460 m/s

c = 0.600 m/s

and that the angle A = 45 degrees = pi/4 radians

Finally then we have

a = sqrt ( 0.46^2 + 0.6^2 - (0.46)(0.6)cos (pi/4))

= sqrt ( 0.5716 - 0.276/sqrt(2))

= sqrt (0.3764) = 0.6135 m/s

The magnitude of the canoe relative to the river is 0.614 m/s. However this is westwards on the East/West axis, so that the canoe is drifting southwards and in the opposite direction to the river on the East/West axis relative to the river. You can tell this from your vector diagram. The precise angle of relative drift can be calculated with the Sine Rule or Cosine Rule.

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