Pre calc word problem

[RPMACS]

A plane takes off 200 miles east of its destination with a 50 mph wind from the south. If the plane’s velocity is 250 mph and its heading is always toward its destination, its path can be described by the equation below. If the location of the initial airport is (0,0) and the location of the destination airport is (200, 0), y represents the perpendicular distance from the straight line path between initial location and the destination location. When x = 150 miles, how far from the straight line path is the plane? That is, what is y?

Y=0.5x^0.8-.5x^12 where 0<or =x<or =200

I really dont know what to do here.

Thanks for your help,

Roger

 

[Deleted member 4993]

A plane takes off 200 miles east of its destination with a 50 mph wind from the south. If the plane’s velocity is 250 mph and its heading is always toward its destination, its path can be described by the equation below. If the location of the initial airport is (0,0) and the location of the destination airport is (200, 0), y represents the perpendicular distance from the straight line path between initial location and the destination location. When x = 150 miles, how far from the straight line path is the plane? That is, what is y?

Y=0.5x^0.8-.5x^12 where 0<or =x<or =200 <<< Is that correct?

I really dont know what to do here.

Thanks for your help,

Roger

 

[galactus]

If the plane is heading due east and the wind is blowing toward the north, then the wind is blowing perpendicular to the plane.

The plane will actually fly \(\displaystyle \sqrt{150^{2}+50^{2}}=50\sqrt{26}\approx 254.95\)

\(\displaystyle tan^{-1}(\frac{50}{250})=tan^{-1}(\frac{1}{5})=11.31 \;\ degrees\) is the bearing to adjust for because of the wind.

If the plane stays on course as if there were no wind, then he will be off course by a little over 11 degrees because of the 50 mph wind blowing against him.

So, there is a triangle to solve. \(\displaystyle 150tan(11.31)=30\)

It will be about 30 miles from the plane to the striaght line path because of the wind.

I reckon I am interpreting right.

 

[RPMACS]

Thank you, this makes worked out