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A commercial airplane that is 1,500 miles into a 2,500-mile journey is traveling at 450 knots in still air when it picks up a tailwind of 150 knots (in the same direction). If h is the number of hours remaining in the airplane's flight, which of the following equations best describes the situation? 1 knot 1.15 miles per hour (mph).
A. 1,500 + 690h = 2,500 00
B. 1,500 + 600h = 2,500
C. 1,500 - 600h = 2,500
D. 1,500 - 690h = 2,500

Respuesta :

oladayoademosu

The equation that best describes the situation is A. 1,500 + 690h = 2,500

Speed of plane after tailwind

Since the plane has already covered 1500 miles, and the plane travelling at 450 knots in still air picks up at tailwind of 150 knots in the same direction, the resultant velocity, V of the plane = speed of plane in still air + speed of wind

= 450 knots + 150 knots

= 600 knots

= 600 × 1 knot

= 600 × 1.15 mph

= 690 mph.

Distance of plane after tailwind

If h is the number of hours after picking up the tail wind, the distance moved by the airplane is thus D = 1500 + 690h

Required equation

Since D = 2500 miles, we have

1500 + 690h = 2500.

So, the equation that best describes the situation is A. 1,500 + 690h = 2,500

Learn more about simple equations here:

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