A jet flies to the west for a distance of approximately 830 miles, starting at Point A and ending at Point B. The jet is moving at 400 mph, on average. A strong wind comes from the north at 40 mph. In minutes, how long would it take for the jet to reach its final destination at Point B?
I'm not quite sure why I'm having issues with this question. I think it might have to do with the part with the wind, unless it's as deceptively simple as doing synthetic division with 400/830?
On
The plane's overall speed is $400$ mph.
Hint: Using the Pythagorean Theorem, if the plane's westward speed is $v$, we know that $v^2+40^2=400^2$
We also know that the plane has to travel $830$ miles WEST.
On
The basic idea is the plane can not be flying due west as the wind would blow it off course. So it must adjust it's heading so that the wind blows it on course.
So the without wind the course and speed of the plane will form the hypotenuse of a right triangle with a magnitude corresponding to $hypotenuse = 400 mph$ and a base angle of $\theta$. The wind forms a leg of the right triangle with a magnitude of $opposite = 40 mph$.
And the actual speed of the plane in a westerly actual over land course will be $adjacent = x$.
So... trig.
$\sin \theta = \frac {40}{400}$ and $\cos \theta = \frac {x}{400}$.
So solve for $x$. Then it will take $\frac {830}x$ hours or $60*\frac {830}x$ minutes to reach the destination.
The plane has to fly in a direction north of west so the wind plus its velocity adds to due westward motion. If you draw a picture you should have a right triangle created by the wind, the plane air speed vector, and the plane ground speed vector. A little trig will see you home.