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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? 1knot=1.15 miles per hour (mph) Choose 1 answer: (A) 1,500+690 h=2,500 (B) 1,500+600 h=2,500 (c) 1,500-600 h=2,500 (D) 1,500-690 h=2,500

A commercial airplane that is 11,500500 miles into a 22,500500 -mile journey is traveling at 450450 knots in still air when it picks up a tailwind of 150150 knots (in the same direction). If h h is the number of hours remaining in the airplane's flight, which of the following equations best describes the situation?\newline11 knot=1.15\mathrm{knot}=1.15 miles per hour (mph)\newlineChoose 11 answer:\newline(A) 1,500+690h=2,500 1,500+690 h=2,500 \newline(B) 1,500+600h=2,500 1,500+600 h=2,500 \newline(C) 1,500−600h=2,500 1,500-600 h=2,500 \newline(D) 1,500−690h=2,500 1,500-690 h=2,500

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Q. A commercial airplane that is 11,500500 miles into a 22,500500 -mile journey is traveling at 450450 knots in still air when it picks up a tailwind of 150150 knots (in the same direction). If h h is the number of hours remaining in the airplane's flight, which of the following equations best describes the situation?\newline11 knot=1.15\mathrm{knot}=1.15 miles per hour (mph)\newlineChoose 11 answer:\newline(A) 1,500+690h=2,500 1,500+690 h=2,500 \newline(B) 1,500+600h=2,500 1,500+600 h=2,500 \newline(C) 1,500−600h=2,500 1,500-600 h=2,500 \newline(D) 1,500−690h=2,500 1,500-690 h=2,500
  1. Calculate Effective Speed: Determine the airplane's effective speed with the tailwind.\newlineThe airplane's speed in still air is 450450 knots, and it picks up a tailwind of 150150 knots. We need to add these two speeds to find the effective speed of the airplane with the tailwind.\newlineEffective speed = 450450 knots + 150150 knots = 600600 knots
  2. Convert to mph: Convert the effective speed from knots to miles per hour (mph). \newline11 knot =1.15= 1.15 mph, so we need to multiply the effective speed in knots by this conversion factor to get the speed in mph. \newlineSpeed in mph =600= 600 knots ×1.15\times 1.15 mph/knot =690= 690 mph
  3. Set Up Equation: Set up the equation to find the remaining hours of flight.\newlineThe airplane has already traveled 1,5001,500 miles and has 1,0001,000 miles left to complete the 2,5002,500-mile journey. The speed of the airplane with the tailwind is 690690 mph. We use the formula distance=speed×time\text{distance} = \text{speed} \times \text{time}, where time is represented by hh (the number of hours remaining).\newline1,5001,500 miles ++ 690690 mph ×h=2,500\times h = 2,500 miles
  4. Verify Equation: Verify that the equation makes sense in the context of the problem.\newlineThe equation 1,500+690h=2,5001,500 + 690h = 2,500 represents the initial distance traveled (1,5001,500 miles) plus the distance that will be covered in the remaining hours of flight (690 mph×h690 \text{ mph} \times h) to equal the total journey distance (2,5002,500 miles). This equation correctly represents the situation described in the problem.\newlineThe equation 1,500+690h=2,5001,500 + 690h = 2,500 matches with the option (A).

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