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MA_775_DIABLO [31]
3 years ago
15

A pilot flies in a straight path for 1 hour 30 minutes. She then makes a course correction, heading 45 degrees to the right of h

er original course, and flies 2 hours 15 minutes in the new direction. If she maintains a constant speed of 345 mi/h, how far is she from her starting position? Give your answer to the nearest mile.
Physics
1 answer:
mrs_skeptik [129]3 years ago
4 0

Answer:

1199 miles

Explanation:

1 hour 30 minutes = 1 + 30/60 = 1.5 hours

2 hours 15 minutes = 2 + 15/60 = 2.25 hours

The distance she flew in the 1st segment is:

1.5*345 = 517.5 miles

The distance she flew in the 2nd segment is:

2.25 * 345 = 776.25 miles

Since the 2nd segment is 45 degree with respect to the 1st segment, this means that she has flown

776.25 * cos(45) = 549 miles in-line with the 1st segment and

776.25* sin(45) = 549 miles perpendicular to the 1st segment:

So the distance from the end to her starting position is

\sqrt{(517.5 + 549)^2 + 549^2} = 1199 miles

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The correct formula to use for the situation given above is: F = MA, where F is the applied force, M is the mass of the object and A is the acceleration.

From the details given in the question, we are told that:

F = 18, 400N

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We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

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Substituting,

W=(11,000)(1600)=17.6\cdot 10^6 J

Now we can calculate the power delivered, which is given by

P=\frac{W}{t}

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1 hp = 746 W

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P=\frac{2.99\cdot 10^4}{746}=40.1 hp

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