A pilot flies in a straight path for 2 h 30 min. She then makes a course correction, heading 7 degrees to the right of her original course, and flies 3h in the new direction. If she maintains a constant speed of 695 mi/h, how far is she from her starting position?
let AB be the original path, then veer off at 7° for BC.
Join AC
I see a long skinny triangle ABC with angle ABC = 173°
The lengths of AB and BC are 2.5(695) and 3(695) respectively, but we can just ignore the 695 for the time being and use the smaller similar triangle to keep the numbers low.
by cosine law
AC^2 = 2.5^2 + 3^2 - 2(2.5)(3)cos173
= 15.25 - (-14.88819...)
= 30.13819..
AC = 5.489826..
so the distance is 5.489..(695) = 3815.4 miles
To find the distance from the starting position, we need to determine the distance covered in each direction.
First, let's convert the time units from hours and minutes to just hours. The pilot flies for 2 hours and 30 minutes, which is equivalent to 2 + 30/60 = 2.5 hours.
She flies at a constant speed of 695 mi/h for 2.5 hours, so the distance covered in the original direction is:
Distance1 = Speed × Time = 695 mi/h × 2.5 h = 1737.5 miles
Next, she makes a course correction and flies for 3 hours at a heading 7 degrees to the right of her original course.
To find the distance in the new direction, we use the cosine function:
Distance2 = Speed × Time × cos(angle)
where the angle is given as 7 degrees. Converting to radians:
Angle = 7 degrees × (π/180) = 7π/180 radians
Now, we can calculate the distance in the new direction:
Distance2 = 695 mi/h × 3 h × cos(7π/180)
Using a calculator to find the cosine value and rounding to the nearest mile:
Distance2 ≈ 695 mi/h × 3 h × 0.9922 ≈ 2065.47 miles ≈ 2065 miles
To find the total distance from the starting position, we sum the distance covered in each direction:
Total distance = Distance1 + Distance2 = 1737.5 miles + 2065 miles = 3802.5 miles
Therefore, the pilot is approximately 3802.5 miles from her starting position.
To solve this problem, we need to break it down into two parts: the distance traveled in the first leg and the distance traveled in the second leg. Then, we can use trigonometry to find the distance from the starting position.
First, let's calculate the distance traveled in the first leg. The pilot flies in a straight path for 2 hours and 30 minutes, which is equivalent to 2.5 hours. Since she maintains a constant speed of 695 mi/h, the distance traveled in the first leg is given by:
Distance1 = Speed1 * Time1 = 695 mi/h * 2.5 h = 1737.5 miles.
Next, let's calculate the distance traveled in the second leg. The pilot makes a course correction, heading 7 degrees to the right of her original course, and flies for 3 hours. To find the distance traveled in this leg, we need to find the horizontal component of this path.
To do this, we can use trigonometric functions. The horizontal component can be found by multiplying the distance traveled in the second leg (Distance2) by the cosine of the angle (7 degrees). Since the pilot maintains a constant speed, the distance traveled in the second leg is given by:
Distance2 = Speed2 * Time2 = 695 mi/h * 3 h = 2085 miles.
Now, we can calculate the horizontal component of the distance:
Horizontal component = Distance2 * cos(angle) = 2085 miles * cos(7 degrees) = 2069.221 miles.
Finally, we can use the Pythagorean theorem to find the distance from the starting position. The distance from the starting position is the hypotenuse of a right triangle, with the vertical component (Distance1) being one side and the horizontal component being the other side. The formula to calculate the distance from the starting position is:
Distance = √(Distance1^2 + Horizontal component^2)
Plugging in the values we've calculated:
Distance = √(1737.5^2 + 2069.221^2) = √(3011525.625 + 4285534.226) = √7297059.851 = 2700.178 miles.
Therefore, the pilot is approximately 2700.178 miles from her starting position.