Question

Help w/ Vector problem?

Hi all, could you please make a drawing and show your work on how you solved this problem?

Answer 1

The speed is `=12.31kmh^-1` in the direction `54.9º` clockwise from the East

Explanation:

You have a triangle `ABC`,

The angle `hat(ABC)=45º`

The speed of the yacht is`AB=10kmh^-1`

The speed of the wind is `BC=3kmh^-1`

The actual speed of the yacht is `AC`

We apply the cosine rule

`AC^2=AB^2+BC^2-2*AB*BC*cos135º`

`AC=sqrt(10^2+3^2+2*10*3*sqrt2/2)`

`=sqrt(109+42.43)`

`=sqrt(151.43)`

`=12.31kmh^-1`

To find the angle `hat(BAC)`, we apply the sine rule

`sin(hat(CAB))/(BC)=sin45/(AC)`

`sin(hat(CAB))=sin45/(AC)*BC`

`=sqrt2/2*3/12.31=0.26`

The angle `hat(CAB)` is `=9.92º`

The direction of the yacht is `=45+9.92=54.9º` clockwise from the East

Answer 2

Actual speed of yacht

`absvec(OC)=sqrt(absvec(OA)^2+absvec(OB)^2+2xxabsvec(OA)xxabsvec(OB)cos45^@)`

`=>absvec(OC)=sqrt(10^2+3^2+2xx10xx3xx1/sqrt2)=12.3kmh^-1`

Applying triangle in `DeltaOBC` we have

`(sin/_BOC)/(BC)=(sin/_OBC)/(OC)`

`=>(sin/_BOC)/10=(sin135^@)/12.3`

`=>sin/_BOC=(10xx1/sqrt2)/12.3`

`/_BOC~~35^@`
So Yacht will travel making `~~35^@`with the south towards east