We have been asked to investigate two flights, AA flight and U A flight. Both flights are at 33,000 feet, and the flight paths i
ntersect directly over city of Frada Heights. A committee of concerned citizens has petitioned the City to investigate the possible danger of a collision of these two flights at 1:30 pm. On the day we checked, we found that at 1:30 PM , AA was 32 nautical miles from Frada Heights, approaching it on a heading of 171 degress at a rate of 405 knots. At the same time, United was 44 nautical miles from Frada Heights and was approaching it on a heading of 81 degrees at a rate of 465 knots. The DPS has asked us to determine how fast the planes were approaching each other (i.e., how fast the distance between them was decreasing) at that time. The DPS would also like to know whether the flights would have violated the FAA's minimum separation requirement of 5 nautical miles. We would like to determine how close the planes actually get to each other.
1 answer:
Answer:
Both flights approach each other at a speed of 624.70 Knots. The FAA minimum separation is not violated as both airplanes are 7.26 Nautical miles away from each other at the time when one of the flights( flight AA) passes through Frada Heights.
Step-by-step explanation:
To solve this kind of problem, the knowledge of concept of relative velocity is needed as the first question requested how fast the flights were approaching each other. To find the minimum distance between both flights, the closest point of approach between both flights should be taken into consideration which was Frada heights. Flight AA passes through Frada Heights in a shorter time of 0.079 hours. This is the time at which both flights approach each other the closest and so the minimum distance (separation) between them. This was calculated to be 7.26NM which is greater than the FAA's minimum this requirement for flight was not violated.
Detailed calculation steps can be found in the attachment below.
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Answer: a) P = 0.5, b) P = 0.07
Step-by-step explanation:
Hi!
Lets call X₁ the time at which you arrive, and X₂ the time at which Bob arrives. Both are random variables with uniform density in the interval [0, 60] (in minutes). Their joint distribuition is uniform over the square in the image, with value P = 1/(60*60) = 1/3600.
a) For you to get more cake than Bob, you should arrive earlier. This event is A = { X₁ < X₂ }, the shaded triangle in the figure.The area of this event (set) is half the total area of the square, so P(A) = 0.5.
It makes sense, beacuse its equally probable for you or Bob to arrive earlier, as both have uniform density over the time interval.
b) In this case you arrive later than Bob, but less than 5 minutes later. So the event is B = { X₂ < X₁ < (X₂ + 5) } . This is the gray shaded area in b) part of the image. Its area is the difference two triangles (half square - blue triangle), then the probability is:
Hi,
Work:
Equation;
1. Write division as fraction.
2. Evaluate the power of 10².
3. Reduce the fraction with 20.
4. Multiply.
5. Finally subtract and RESULT.
Hope this helps.
r3t40
Work:
Equation;
1. Write division as fraction.
2. Evaluate the power of 10².
3. Reduce the fraction with 20.
4. Multiply.
5. Finally subtract and RESULT.
Hope this helps.
r3t40