Answer to Question #94085 in Mechanics | Relativity for Bagdat

Question

A ship moves along the equator to the east with velocity v=30 kin/hour.The southeastern wind blows at an angle 60 to the equator with velocity v=15 km/hour.Find the wind velocity v relative to the ship and the angle between the equator and the wind direction in reference frame fixed to the ship.

Answer 1

$\overrightarrow{V_0}$ - velocity of the boat relative to the equator

$\overrightarrow{V_w}$ - velocity of wind relative to the equator

$\overrightarrow{V_r}$ - velocity of wind relative to the ship

$\alpha$ - angle between the equator and the wind direction in reference frame fixed to the ship

According to law of adding velocities, velocity of wind relative to the equator is the vector sum of 2 vectors: velocity of the boat relative to the equator and velocity of wind relative to the boat.

$\overrightarrow{V_w}=\overrightarrow{V_0}+\overrightarrow{V_r}$

According to the rule of adding vectors, we can draw the following picture:

According to cosine rule,

$|\overrightarrow{V_r}|^2={|\overrightarrow{V_w}|^2+|\overrightarrow{V_0}|^2-2|\overrightarrow{V_w}||\overrightarrow{V_0}|cos120^o}$

$|\overrightarrow{V_r}|=\sqrt{|\overrightarrow{V_w}|^2+|\overrightarrow{V_0}|^2-2|\overrightarrow{V_w}||\overrightarrow{V_0}|cos120^o}$

$|\overrightarrow{V_r}|=\sqrt{15^2+30^2-2*15*30*(-0.5)}$

$|\overrightarrow{V_r}|=\sqrt{225+900+450}=\sqrt{1575}=15\sqrt{7} \dfrac{km}{h}\approx39.7\dfrac{km}{h}$

According to cosine rule,

$|\overrightarrow{V_w}|^2={|\overrightarrow{V_r}|^2+|\overrightarrow{V_0}|^2-2|\overrightarrow{V_r}||\overrightarrow{V_0}|cos\alpha}$

$|\overrightarrow{V_w}|^2-|\overrightarrow{V_r}|^2-|\overrightarrow{V_0}|^2=-2|\overrightarrow{V_r}||\overrightarrow{V_0}|cos\alpha$

$|\overrightarrow{V_r}|^2+|\overrightarrow{V_0}|^2-|\overrightarrow{V_w}|^2=2|\overrightarrow{V_r}||\overrightarrow{V_0}|cos\alpha$

$\dfrac{|\overrightarrow{V_r}|^2+|\overrightarrow{V_0}|^2-|\overrightarrow{V_w}|^2}{2|\overrightarrow{V_r}||\overrightarrow{V_0}|}=cos\alpha$

$cos\alpha=\dfrac{|\overrightarrow{V_r}|^2+|\overrightarrow{V_0}|^2-|\overrightarrow{V_w}|^2}{2|\overrightarrow{V_r}||\overrightarrow{V_0}|}=\dfrac{1575+900-225}{2*15\sqrt{7}*30}$

$cos\alpha=\dfrac{1575+900-225}{2*15\sqrt{7}*30}=\dfrac{2250}{900\sqrt{7}}=\dfrac{5}{2\sqrt{7}}$

$\alpha=arccos(\dfrac{5}{2\sqrt{7}})\approx19.1^o$

Answer: $|\overrightarrow{V_r}|=15\sqrt{7} \dfrac{km}{h}\approx39.7\dfrac{km}{h}; \alpha=arccos(\dfrac{5}{2\sqrt{7}})\approx19.1^o$