Question

An airplane, flying at a constant speed of 360 mi/hr and climbing at a 30 degree angle, passes over a point P on the ground at an altitude of 21,120 ft. Find the rate (in miles per hour) at which its distance from P is changing one minute later.

Answer 1

Answer:

Step-by-step explanation:

Given  

velocity of airplane is 360 mi/hr

inclination of climbing $30^{\circ}$

Plane velocity can be divided in to x & y component

Velocity in x component is $360\cos30$

Velocity in y component is $360\sin30$

Thus distance traveled in x axis is $=360\cos30\times \frac{1}{60}$

=60cos30=51.96 miles

Distance travel in upward direction i.e. in Y axis

$=360\sin30\times \frac{1}{60}=30 miles$

so total altitude gain by plane is 30+4=34 miles

Horizontal distance traveled is 51.96

inclination w.r.t to point p

$tan\theta =\frac{34}{51.96}=0.654$

$\theta =33.18^{\circ}$

Initial inclination is $90^{\circ}$

therefore rate at which its distance is changing is $360\cos30=311.76 miles /hour$