A speedboat leaves port pointed due East at 80 knots (150 km/h). The ocean in this region has a northward current of 4.00m/s

Question

A speedboat leaves port pointed due East at 80 knots (150 km/h). The ocean in this region has a northward current of 4.00m/s

a. What are the components of the boat’s velocity vector relative to the starting port? What is the angle of travel? b. If the skipper intended for the boat to go due east, how far off course is the boat after driving like this for 30 min?

c. How far, in total has the boat travelled in that time?

Answer 1

A speedboat leaves port pointed due East at 80 knots (150 km/h). The ocean in this region has a northward current of 4.00m/s

a. What are the components of the boat’s velocity vector relative to the starting port? What is the angle of travel?

First, convert all aspects of the question to consistent units, m/s is the basic unit of velocity in the metric system (MKS): 150 km/h = 150,000 m/hr / 3600 sec/hr = 41.7 m/s. Next step, using basic trig functions (SohCahToa), describe V (the resultant of both vectors, the 4 m/s of the northward current V(y) and the boat's initial eastward vector V(x): V(y) = V*sin (alpha) where alpha is the angle between both vectors, V(x) = V*cos alpha. V(x) = 41.7 m/s, V(y) = 4 m/s. Note, we've already determined V(x) and V(y) and didn't need to use any trig identities for this step but, in already doing so, we can now compute the angle using the trig identity for the Tangent: alpha = arctan (4/41.7) = 5.48 degrees, an angle that is consistent with the gentle northward current and the boat's eastward velocity of 42 m/s.

Using the Pythagorean theorem, we compute the resultant true velocity of the boat as : V(x)² + V(y)² = V², V = sqrt (16 + 1738.9) = 41.9 m/s, again a result that is consistent with the gentle northward current.

b. If the skipper intended for the boat to go due east, how far off course is the boat after driving like this for 30 min? To determine this, we convert 30 min to seconds: (1800) and then multiply that by V(y): 4m/s * 1800s = 7200 m or 7.2 km NE of where they expected to be after 30 minutes. The total path length traveled by the boat (the resultant velocity * time = length of the resultant vector in meters: 41.9 m/s * 1800 seconds) = 41.9 m/s * 1800 s = 75420 m = 75.42 km.

c. How far, in total has the boat traveled in that time? From above: 75420 m or 75.42 km.

Answer 2

a. The boat has an ( x ) component of motion. It has no ( y ) component of its own, but the ocean current moves in the ( y ) direction. For this reason, we must add these velocities to determine the 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 ) = sqrt [ ( 41.7 m / s )^2 + ( 4.00 m / s )^2 ] = 41.9 m / s North of East.

Since tan ( theta ) = ( opp / adj ), tan ( theta ) = [ ( 4.00 m / s ) / ( 41.7 m / s ) ]. Therefore, tan^-1 [ 0.0959 ] = 5.48 degrees North of East.

b. 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 whose angle North of East is 5.48 degrees. The boat will miss its Eastern point by a distance equal to sin 5.48 = ( opp / 75.4 km ). Therefore, opp = ( 75.4 km )( sin 5.48 ) = 7.20 km.

c. The boat has traveled 75.4 km.