Answer to Question #169817 in Classical Mechanics for Akin

Question #169817

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


1
Expert's answer
2021-03-08T08:15:58-0500

Let vector aa represents the displacement 2.0 km east and vector bb represents the displacement 3.50 km southeast. Let's find the resultant displacement:


R=a+b,R=a+b,R=(2.0 km,0)+(3.5 kmcos45,3.5 kmsin45),R=(2.0\ km, 0)+(3.5\ km\cdot cos45^{\circ}, -3.5\ km\cdot sin45^{\circ}),R=(2.0 km+3.5 kmcos45)i^+(03.5 kmsin45)j^,R=(2.0\ km+3.5\ km\cdot cos45^{\circ})\hat{i}+(0-3.5\ km\cdot sin45^{\circ})\hat{j},R=4.47i^2.47j^.R=4.47\hat{i}-2.47\hat{j}.

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


R=Rx2+Ry2=(4.47 km)2+(2.47 km)2=5.11 km.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:


θ=sin1(RyR)=sin1(2.47 km5.11 km)=28.9.\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 S of E28.9^{\circ}\ S\ of\ E.




Now, let vector aa represents the displacement 5.11 km 28.9 S of E28.9^{\circ}\ S\ of\ E, vector bb represents the unmeasured displacement which we are searching for, and vector RR 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,R=a+b,b=Ra,b=R-a,b=(5.8 km,0)(5.11 kmcos28.9,5.11 kmsin28.9),b=(5.8\ km, 0)-(5.11\ km\cdot cos28.9^{\circ}, 5.11\ km\cdot sin28.9^{\circ}),b=(5.8 km5.11 kmcos28.9)i^(05.11 kmsin28.9)j^,b=(5.8\ km-5.11\ km\cdot cos28.9^{\circ})\hat{i}-(0-5.11\ km\cdot sin28.9^{\circ})\hat{j},b=(1.33i^+2.47j^).b=(1.33\hat{i}+2.47\hat{j}).

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


R=Rx2+Ry2=(1.33 km)2+(2.47 km)2=2.8 km.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:


θ=sin1(RyR)=sin1(2.47 km2.8 km)=62.\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 62N of E.62^{\circ} N\ of\ E.

Therefore, the unmeasured displacement of the sailor has magnitude 2.8 km and direction 62N of E62^{\circ} N\ of\ E.


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