Question

A man is walking on a ship as the ship approaches the shore. If the man's velocity is 1.80 m/s due north relative to the ship, and 4.50 m/s at an angle of 31.0° west of north relative to the shore, what are the direction and magnitude of the ship's velocity relative to the shore?

Answer 1

Let's call A the man, B the ship and C the shore.

The data we have it's all the veolicity relative to some point:

VAB = Velocity relative from man to ship = 1.8 m/s

VAC: Velocity relative from man to the shore = 4.50; angle: 31 °

VBC: Velocity relative from ship to the shore = Unknown

So, let's write the equations:

VAB = VA - VB (1)

VAC = VA - VC (2)   

VBC = VB - VC (3) --------> From here: VC = VB - VBC   (4)

Now, let's substitute (4) in (2):

VAC = VA - VB - VBC -----> but VA - VB = VAB so:

VAC = VAB - VBC and solving for VBC:

VBC = VAB - VAC (5)

This would be the equation to use to calculate the velocity of the ship relative to the shore:

VAC_x = 4.5 cos31 = 3.86 m/s

VAC_y = 4.5 sin31 = 2.32 m/s

VBC_x = 3.86 m/s

VBC_y = 1.8 - 2.32 = -0.52 m/s

|VBC| = [(0.52²) + (3.86²)]^1/2

|VBC| = 3.89 m/s

Tan A = 0.52/3.86

Tan A = 0.1347

A = 7.67 °