Question

A plane has an airspeed of 400 mph . The pilot wishes to reach a destination 800 mi due east, but a wind is blowing at 60 mph in the direction 30 ∘ north of east.
Part A
In what direction must the pilot head the plane in order to reach her destination?
Part B
How long will the trip take?

Answer 1

The plane's heading is 4.61 degrees south of east, which is the direction the pilot needs to steer the plane in order to reach her objective and the trip takes 1.83 hours.

What is the distance?

Distance is a numerical representation of the distance between two objects or locations.

The distance can refer to a physical length or an estimate based on other factors in physics or common use. |AB| is a symbol for the distance between two points A and B.

Let the pilot's heading be theta south with an east direction. The resultant the velocity will be;

$\rm 400 sin(\theta) - 50 sin(40) = 0 \\\\\ \rm 400 sin(\theta) - 37.25 = 0 \\\\ \ \theta= sin^{-1}\frac{37.25}{400}\\\\ \theta = 4.61^0$

Hence the direction of the plane heading is 4.61 degrees south of east

b)

The resultant speed of the plane(V)is found as;

$\rm V= 400 cos(4.61) + 50 cos(40)\\\\ V = 437 miles/hr \\\\ T = \frac{800}{437}\\\\ T = 1.83 hour$

Hence, the trip takes 1.83 hours.

To learn more about the distance refer to the link;

brainly.com/question/26711747

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