Velocity Q?

Question

A boat heads 15˚ west of north with a water speed of 3m/s. Determine its velocity relative to the ground when there is a 2 m's current from 40˚ east of north?

Answer 1

You have to do some trigonometry with this one.  First lets look at the motion of the boat.  If you quickly sketch a line of 3 m/s at 15° W of N, draw a right angle down from the tip of your 3 m/s vector line.  The inside angle will be 90-15 = 75°.  Sin75 * 3 = 2.89 m/s of vertical velocity.  Cos 75 * 3 = 0.77 m/s [West] horizontal velocity.

Now for the motion of the water.  Draw another triangle as above.  The water is moving from NE so you have to draw the motion of the water as heading SW.  Sin40 * 2 = 1.28 m/s [West] horizontal velocity.  Cos 40 * 2 = 1.53 m/s [South] vertical velocity.

Add the two horizontal and vertical velocities together and draw a new triangle with that info.  South and West are negatives, North and East are positives.  

Horizontal:  1.28 m/s [West] + 0.77 m/s [West] = 2.05 m/s [West].

Vertical:  2.89 m/s [North] -1.53 m/s [South] = 1.36 m/s [North].

With your new triangle you can solve for the angle relative to the ground by taking the tangent (opposite/adjacent) of the sides.  Tan 1.3/2.05 = 32.4 °.  From your diagram it should be clear that this direction is 32.4° North of West.  

Use Pythagorean to solve for the hypotenuse.  1.3 squared + 2.05 squared = 5.8925.  Square root of hypotenuse is 2.43.

Final answer is 2.43 m/s at 32° North of West, or 58° West of North.