frames of reference/vectors question.

[seang]

The captain of a boat wants to travel directly across a river that flows due east with a speed of 1.48 m/s. He starts from the south bank of the river and wants to reach the north bank by travelling straight across the river. The boat has a speed of 5.64 m/s with respect to the water. What direction (in degrees) should the captain steer the boat? Note that 90° is east, 180° is south, 270° is west, and 360° is north.

In this problem, we were told to find the answer by taking 360 minus the inverse tangent of windspeed/boatspeed. I don't understand why the 'boatspeed with respect to the water' should be perpendicular to the shore. It seems to me that since the water is moving, the the 5.64m/s with respect to the water should be the hypotenuse of the triangle.

What's wrong with my thinking?

[Dorothy Weglend]

In this problem, we were told to find the answer by taking 360 minus the inverse tangent of windspeed/boatspeed. I don't understand why the 'boatspeed with respect to the water' should be perpendicular to the shore. It seems to me that since the water is moving, the the 5.64m/s with respect to the water should be the hypotenuse of the triangle.

What's wrong with my thinking?


The downstream component of the velocity is perpendicular to the shore because the river is flowing upstream, also perpendicular to the shore. To go straight across, the boat's velocity needs a downstream component to cancel the upstream 'push' the river would give it.

The hypotenuse of the triangle, in this case, would be the straight line velocity that would be followed by the boat in still water. The other leg is the vector sum of these two vectors.

Dot

[seang]

wait, why isn't the downstream component of the velocity parallel to shore?

[Dorothy Weglend]

wait, why isn't the downstream component of the velocity parallel to shore?

Yes. I'm sorry. The downstream component is parallel to the shore, perpendicular to the path actually travelled by the boat.

I think I need coffee.

Dot