Thread: boat in water, force vectors

A boat heads in the direction N 40 degrees W at a speed of 24 knots. The current runs at 6 knots in the direction N 25 degrees E. Find the resultant speed and direction of the boat.

Hello, Mr_Green!

I must assume you are familiar with vectors.

A boat heads in the direction N 40° W at a speed of 24 knots.
The current runs at 6 knots in the direction N 25° E.
Find the resultant speed and direction of the boat.

Code:

            E  6·sin25° C
            + - - - - - *
            :         /
   6·cos25° :       / 6
            :     /
            :25°/
            : /  24·sin24°
          B * - - - - - - + D
              \           :
                \         :
                  \       : 24·cos24°
                24  \     :
                      \40°:
                        \ :
                          *
                          A

The boat is moving from point A to point B.
. . Its horizontal component is: DB = -24·sin24°
. . Its vertical component is: AD = 24·cos24°

The current is moving from point B to point C.
. . Its horizontal component is: EC = 6·sin25°
. . Its vertical component is: BE = 6·cos25°

The resultant vector is AC.
We must determine its magnitude and direction.


The horizontal component of AC is: .-24·sin24° + 6·sin25° .≈ .-7.226
The vertical component of AC is: .24·cos24° + 6·cos25° .≈ .27.363
. . . . . . . . . . . . . . . . . . . . . ________________
The magnitude is: .|AC| .= .√(-7.226)² + 27.363² .= .28.30103965

. . Therefore, the speed of the boat is about 28.3 knots.


The direction angle is given by: .tanθ .= .7.226/27.363 . → . θ .= .14.79292136°

. . Therefore, the direction is about: .N 14.8° W