Respuesta :
Answer:
Step-by-step explanation:
Given
velocity of airplane is 360 mi/hr
inclination of climbing [tex]30^{\circ}[/tex]
Plane velocity can be divided in to x & y component
Velocity in x component is [tex]360\cos30 [/tex]
Velocity in y component is [tex]360\sin30 [/tex]
Thus distance traveled in x axis is [tex]=360\cos30\times \frac{1}{60}[/tex]
=60cos30=51.96 miles
Distance travel in upward direction i.e. in Y axis
[tex]=360\sin30\times \frac{1}{60}=30 miles [/tex]
so total altitude gain by plane is 30+4=34 miles
Horizontal distance traveled is 51.96
inclination w.r.t to point p
[tex]tan\theta =\frac{34}{51.96}=0.654[/tex]
[tex]\theta =33.18^{\circ}[/tex]
Initial inclination is [tex]90^{\circ}[/tex]
therefore rate at which its distance is changing is [tex]360\cos30=311.76 miles /hour[/tex]
The point P is a point below the path of travel of the plane such as the
location of a house close to an airport.
Rate of change of the distance of the airplane from P is [tex]\underline{\dfrac{1440\cdot \sqrt{19} }{19} \ \dfrac{km}{hr}}[/tex]
Reason:
The given parameters are;
The speed of the airplane = 360 mi/hr
The degree angle of the airplane = 30°
The height from which the airplane passes the point, P = 21,120 ft.
The distance the airplane travels per minute from the point in the x-direction is given as follows;
- [tex]D_x = \dfrac{360}{60} \times t \times cos (30^{\circ}) = 6 \cdot t \cdot cos (30^{\circ})[/tex]
The distance the airplane travels per minute from the point in the y-direction is given as follows;
- [tex]D_y = 4 + \dfrac{360}{60} \times t \times sin(30^{\circ}) = 4 + 6 \cdot t \cdot sin(30^{\circ})[/tex]
The magnitude of the distance, D, is given, by Pythagoras's theorem, as follows;
- [tex]D = \sqrt{D_x^2 + D_y^2}[/tex]
- [tex]D^2 = D_x^2 + D_y^2}[/tex]
Therefore;
D² = (6·t·cos(30°))² + (4 + 6·t·sin(30°))²
D² = 36·t²·cos²(30°) + 16 + 24·t·sin(30°) + 36·t²·sin²(30°)
D² = 36·t²·cos²(30°) + 36·t²·sin²(30°) + 16 + 24·t·sin(30°)
D² = 36·t²·(cos²(30°) + sin²(30°)) + 16 + 24·t·sin(30°)
- D² = 36·t² + 16 + 24·t·sin(30°) = 36·t² + 24·t + 16
[tex]\dfrac{d}{dt} D^2 = \dfrac{d}{dt} (36 \cdot t^2 + 24 \cdot t + 16)[/tex]
[tex]2 \cdot D \cdot\dfrac{d}{dt} D = 72\cdot t + 24[/tex]
- [tex]\dfrac{dD}{dt} = \dfrac{72\cdot t + 24}{2 \cdot D}[/tex]
The distance covered in a minute, is given as follows;
D = √(6 × 1 × cos(30°))² + (4 + 6 × 1 × sin(30°))²) = 2·√19
The rate of change of the distance in a minute is therefore;
t = 1
[tex]\dfrac{dD}{dt} = \dfrac{72\times 1 + 24}{2 \times 2 \times \sqrt{19} } = \dfrac{24 \cdot \sqrt{19} }{19}[/tex]
- [tex]\dfrac{dD}{dt} = \dfrac{24 \cdot \sqrt{19} }{19} \ km/min[/tex]
The rate of change of distance in miles per hour is found as follows;
- [tex]\dfrac{dD}{dt} \ in \ mi/hr = 60 \ min/hr\times \dfrac{24 \cdot \sqrt{19} }{19} \ km/min = \dfrac{60\cdot24 \cdot \sqrt{19} }{19} \ km/hr[/tex]
[tex]\dfrac{dD}{dt} \ in \ mi/hr = \dfrac{60\cdot24 \cdot \sqrt{19} }{19} \ km/hr = \dfrac{1440\cdot \sqrt{19} }{19} \ km/hr[/tex]
The rate at which its distance from P is changing one minute later is therefore;
- [tex]\underline{\dfrac{1440\cdot \sqrt{19} }{19} \ \dfrac{km}{hr}}[/tex]
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