A speedboat travels eastward at 80 knots (150 km/h) through a northward current of water moving 4.00m/s.
Q1: What is the speedboat’s velocity vector relative to the starting port? What is the angle of travel?
Q2: If the boat’s captain wanted the boat to travel due east without getting off course, how far off course is the boat after traveling through the northward moving current for 30 min?
Q3: How far, in total, has the speedboat traveled during this period of time?
A1: The speedboat has an x-component of motion. It has no y-component of its own, but the northward current of water moves in the ( +y ) direction. For this reason, we must add these velocities to determine the speedboat’s net direction of motion. First, however, we must convert the boat’s velocity to metric units:
( 150 km / h )( 1 h / 3,600 s )( 1,000 m / 1 km ) = 41.7 m/s east.
Using the Pythagorean theorem, the velocity of the boat ( v ) = √ [ ( 41.7 m/s )2 + ( 4.00 m/s )2 ] = 41.9 m/s in a northeastward direction.
Since tan ( θ ) = ( opp / adj ), tan ( θ ) = [ ( 4.00 m/s ) / ( 41.7 m/s ) ]. Therefore, tan-1 [ 0.0959 ] = 5.48 degrees northeast.
A2: The distance traveled will be ( 41.9 m/s )( 1,800 s ) = 75,420 m = 75.4 km. This will form the hypotenuse of a right triangle with a northeast angle of 5.48 degrees relative to the intended eastward direction of travel; therefore, the boat will miss its due east destination by a distance equal to the opposite side of the right triangle. Thus, sin ( 5.48 degrees ) = ( opp / 75.4 km ). Therefore, opp = ( 75.4 km )( sin 5.48 degrees ) = 7.20 km.
A3: The boat has traveled 75.4 km.
RENAISSANCE PHYSICS 