Answer to Question #169033 in Physics for Shuta

Question

A sailor in a small sailboat encounters

shifting winds. She sails 2.00 km east, then

3.50 km southeast, and then an additional

distance in an unknown direction. Her final

position is 5.80 km directly east of the starting

point. Find the magnitude and direction of the

third leg of the journey. Draw the vector

addition diagram and show that it is in

qualitative agreement with your numerical

solution

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 $62^{\circ} N\ of\ E$.