The actual speed or velocity of the man relative to the earth is 15.52km/hr.
The direction of the man's velocity relation to earth is 75° South of West.
Given the data in the question;
- Velocity of man relative to ship; [tex]v_{mrs} = 4 km/hr[/tex] west
- Velocity of ship; [tex]v_s = 15 km/hr[/tex] south
- Velocity of the man relative to the earth; [tex]v_E =\ ?[/tex]
- Direction of the man's velocity; [tex]\theta = \ ?[/tex]
To get the velocity of the man relative to the earth. From Pythagorean theorem
We can simply say
[tex]v_E = \sqrt{(v_{mrs})^2 + (v_s)^2}[/tex]
We substitute in our given values
[tex]v_E = \sqrt{(4km/hr)^2 + (15km/hr)^2}\\\\v_E = \sqrt{(16km^2/hr^2) + (225km^2/hr^2})\\\\v_E = \sqrt{241km^2/hr^2}\\\\v_E = 15.52 km/hr[/tex]
Therefore, the actual speed or velocity of the man relative to the earth is 15.52km/hr
To get the direction of the man's speed. From the three main trigonometric ratios: SOH CAH TOA
[tex]tan\theta = \frac{opposite}{adjacent}[/tex]
[tex]tan\theta = \frac{v_s}{v_{mrs}}[/tex]
We substitute in our values;
[tex]tan\theta = \frac{15km/hr}{4km/hr}\\\\tan\theta = 3.75\\\\\theta = tan^{-1}( 3.75)\\\\\theta = 75^o[/tex]
Therefore, the direction of the man's velocity relation to earth is 75° South of West
Learn more; https://brainly.com/question/5046811
