Answer to Question #167144 in Physics for Arvie Romana Leal

Question

A traveller is riding a boat. She sails 3.50 km east, then 2.00 km southeast, and then an additional

unmeasured displacement. Her final position is 6.90 km directly east of the starting point. Calculate the

unmeasured displacement. (a) Use method of triangle (b) Use addition of vector components

Answer 1

(a) Let vector $a$ represents the displacement 3.50 km east and vector $b$ represents the displacement 2.00 km southeast. Let's find the resultant displacement:

$$R=a+b,$$ $$R=(3.5\ km, 0)+(2.0\ km\cdot cos45^{\circ}, -2.0\ km\cdot sin45^{\circ}),$$ $$R=(3.5\ km+2.0\ km\cdot cos45^{\circ})\hat{i}+(0-2.0\ km\cdot sin45^{\circ})\hat{j},$$ $$R=4.91\hat{i}-1.41\hat{j}.$$

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

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{(4.91\ km)^2+(-1.41\ km)^2}=5.11\ km.$$

We can find the direction from the geometry:

$$\theta=sin^{-1}(\dfrac{R_y}{R})=sin^{-1}(\dfrac{-1.41\ km}{5.11\ km})=-16^{\circ}.$$

The sign minus means that the resultant displacement has direction $16^{\circ}\ S\ of\ E$.

Now, let vector $a$ represents the displacement 5.11 km $16^{\circ}\ S\ of\ E$, vector $b$ represents the unmeasured displacement which we are searching for, and vector $R$ represents the final resultant displacement of the boat traveller -  6.90 km directly east of the starting point.

Let's find the unmeasured displacement:

$$R=a+b,$$ $$b=R-a,$$ $$b=(6.9\ km, 0)-(5.11\ km\cdot cos16^{\circ}, 5.11\ km\cdot sin16^{\circ}),$$ $$b=(6.9\ km-5.11\ km\cdot cos16^{\circ})\hat{i}-(0-5.11\ km\cdot sin16^{\circ})\hat{j},$$ $$b=(1.98\hat{i}+1.41\hat{j}).$$

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

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{(1.98\ km)^2+(1.41\ km)^2}=2.43\ km.$$

We can find the direction from the geometry:

$$\theta=sin^{-1}(\dfrac{R_y}{R})=sin^{-1}(\dfrac{1.41\ km}{2.43\ km})=35^{\circ}.$$

The sign plus means that the unmeasured displacement has direction $35^{\circ} N\ of\ E.$

(b) Let's find the unmeasured displacement by addition of vector components.

$$R_x=a_x+b_x+x,$$ $$6.9\ km=3.5\ km+2.0\ km\cdot cos45^{\circ}+x,$$ $$x=6.9\ km-3.5\ km-2.0\ km\cdot cos45^{\circ}=1.98\ km.$$ $$R_y=a_y+b_y+y,$$ $$0=0+(-2.0\ km\cdot sin45^{\circ})+y,$$ $$y=2.0\ km\cdot sin45^{\circ}=1.41\ km.$$

Therefore, we get:

$$b=1.98\hat{i}+1.41\hat{j}.$$

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

$$R=\sqrt{R_x^2+R_y^2}=\sqrt{(1.98\ km)^2+(1.41\ km)^2}=2.43\ km.$$

We can find the direction from the geometry:

$$\theta=sin^{-1}(\dfrac{R_y}{R})=sin^{-1}(\dfrac{1.41\ km}{2.43\ km})=35^{\circ}\ N\ of\ E.$$