Answer to Question #169817 in Classical Mechanics for Akin

Question

A sailor in a small sail boat encounter shifting winds. She sails 2.00km East , then 3.50km 45° Southeast, and then an additional distance in an unknown direction

Answer 1

Let vector $a$ represents the displacement 2.0 km east and vector $b$ represents the displacement 3.50 km southeast. Let's find the resultant displacement:

$$R=a+b,$$ $$R=(2.0\ km, 0)+(3.5\ km\cdot cos45^{\circ}, -3.5\ km\cdot sin45^{\circ}),$$ $$R=(2.0\ km+3.5\ km\cdot cos45^{\circ})\hat{i}+(0-3.5\ km\cdot sin45^{\circ})\hat{j},$$ $$R=4.47\hat{i}-2.47\hat{j}.$$

We can find the magnitude of the resultant displacement from the Pythagorean theorem:

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{(4.47\ km)^2+(-2.47\ km)^2}=5.11\ km.$$

We can find the direction from the geometry:

$$\theta=sin^{-1}(\dfrac{R_y}{R})=sin^{-1}(\dfrac{-2.47\ km}{5.11\ km})=-28.9^{\circ}.$$

The sign minus means that the resultant displacement has direction $28.9^{\circ}\ S\ of\ E$.

Now, let vector $a$ represents the displacement 5.11 km $28.9^{\circ}\ S\ of\ E$, vector $b$ represents the unmeasured displacement which we are searching for, and vector $R$ represents the final resultant displacement of the sailor - 5.80 km directly east of the starting point.

Let's find the unmeasured displacement:

$$R=a+b,$$ $$b=R-a,$$ $$b=(5.8\ km, 0)-(5.11\ km\cdot cos28.9^{\circ}, 5.11\ km\cdot sin28.9^{\circ}),$$ $$b=(5.8\ km-5.11\ km\cdot cos28.9^{\circ})\hat{i}-(0-5.11\ km\cdot sin28.9^{\circ})\hat{j},$$ $$b=(1.33\hat{i}+2.47\hat{j}).$$

We can find the magnitude of the resultant displacement from the Pythagorean theorem:

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{(1.33\ km)^2+(2.47\ km)^2}=2.8\ km.$$

We can find the direction from the geometry:

$$\theta=sin^{-1}(\dfrac{R_y}{R})=sin^{-1}(\dfrac{2.47\ km}{2.8\ km})=62^{\circ}.$$

The sign plus means that the unmeasured displacement has direction $62^{\circ} N\ of\ E.$

Therefore, the unmeasured displacement of the sailor has magnitude 2.8 km and direction $62^{\circ} N\ of\ E$.