Question

Kinematics Question Please Help.

A pilot is flying from Montreal, a displacement of 571 km [E 13.0(degrees) N]. The airplane is capable of flying at 301 km/h in still air. During the flight there is a constant wind of 95 km/h blowing form the direction of [E 30.0(degrees) N]. Assuming she flies directly to Montreal, (along the displacement vector) calculate the plane's speed relative to the ground and the direction the pilot should point the airplane to maintain his direct route to Montreal.

Answer 1

The velocity of the air relative to the ground is

`\vec v_(A/G)=\left(95\,(\rm km)/(\rm h)\right)(\cos30.0^\circ\,\vec\imath+\sin30.0^\circ\,\vec\jmath)`

The velocity of the plane relative to the air is

`\vec v_(P/A)=\left(301\,(\rm km)/(\rm h)\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)`

where
`\theta` is the direction the plane needs to point so that the resultant vector - the velocity of the plane relative to the ground - is

`\vec v_(P/G)=\vec v_(P/A)+\vec v_(A/G)=v(\cos13.0^\circ\,\vec\imath+\sin13.0^\circ\,\vec\jmath)`

Here
`v` is the speed of the plane relative to the ground.

So we have

`\begin{cases}v\cos13.0^\circ=\left(95\,(\rm km)/(\rm h)\right)\cos30.0^\circ+\left(301\,(\rm km)/(\rm h)\right)\cos\theta\\v\sin13.0^\circ=\left(95\,(\rm km)/(\rm h)\right)\sin30.0^\circ+\left(301\,(\rm km)/(\rm h)\right)\sin\theta\end{cases}`

Use a calculator to solve for
`v` and
`\theta`; you should find

`v=390\,(\rm km)/(\rm h)`

`\theta=7.7^\circ`