Question

Help w/ another Vector problem?

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

Answer 1

The airplane must head `2.86º` clockwise from the East
The actual speed is `=801kmh^-1`

Explanation:

We have a right angle triangle `ABC`

The speed of the plane is `AB=800kmh^-1`

The speed of the wind is `BC=40kmh^-1`

The angle `hat(ABC)=90º`

We apply Pythagoras' theorem

`AC^2=AB^2+BC^2`

`=800^2+40^2=641600`

So,

`AC=sqrt(641600)=801`

The actual speed is `=801kmh^-1`

To calculate the direction,

`tan(hat(BAC))=(BC)/(AC)`

`=40/800=0.05`

angle `hat(BAC)=2.86º`

The airplane must head `2.86º` clockwise from the East

Answer 2

The pilot should fly 204 km/h at an angle of `11^@` North East.

Explanation:

First, let's look at a drawing of the word problem:

The pilot wishes to go straight east. So she wants the resultant vector to be `vec (OA)`. However, because of the wind, she must fly along vector `vec (BA)`.

`vec (BA)= vec (OA)-vec (OB)`

Luckily, we have a right triangle and `vec (BA)` should be easy to solve. The Pythagorean Theorem tells us that

`|BA|^2=|OB|^2+|OA|^2`
`|BA|^2=40^2+200^2=41600`
`|BA|=204 " km"//"h"`

To calculate the direction, lets look at the angle `theta=/_(OBA)`

`sin(theta)="opp"/"hyp"=(|OA|)/(|BA|)`
`sin(theta)=200/204`
`theta=sin^(-1)(200/204)=79^@`

From the origin, that angle would be `90^@-79^@=11^@`