The pilot of a small plane finds that the airport where he intended to land is fogged in. He flies 55 miles west to another airport to find that conditions there are too icy for him to land. He flies 25 miles at 15 degrees east of south and is finally able to land at the third airport. (a) how far and in what direction must he fly the next day to go directly to his original destination? (b) how many extra miles beyond his original flight plan has he flown?
(a) Add the vectors of his 55 mile and 25 miles flights. The reverse of that vector is the direction and distance he must fly the next day.
(b) Add the magnitudes of the three flight vectors in the previous question.
To solve this problem, we can use vector addition and trigonometry.
Let's break down the pilot's flight into different legs:
1. Leg 1: The pilot flies 55 miles west from the original destination.
2. Leg 2: The pilot flies 25 miles at a 15-degree angle east of south.
3. Leg 3: Unknown distance and direction to reach the original destination.
Now let's solve the problem step-by-step:
(a) To find the distance and direction the pilot needs to fly the next day to reach the original destination directly, we need to add all the vectors.
First, let's represent the vectors graphically:
N
|
55 miles (west)
|
W <---Pilot---> E
\
25 miles
\
S
From the diagram, we can see that the pilot needs to fly back east to reach the original destination.
To find the distance, we need to calculate the resultant vector's magnitude (hypotenuse):
Resultant magnitude = √[(Leg1)^2 + (Leg2)^2]
= √[(55 miles)^2 + (25 miles)^2]
= √[3025 + 625]
= √3650
≈ 60.42 miles
The pilot needs to fly approximately 60.42 miles.
To find the direction, we can use trigonometry. Since we started with a 15-degree angle east of south, the pilot needs to fly the opposite direction, which is 180 degrees - 15 degrees = 165 degrees.
So, the pilot needs to fly approximately 60.42 miles in a direction of 165 degrees.
(b) To find how many extra miles beyond the original flight plan the pilot has flown, we need to compare the total distance he has flown to the original distance.
The total distance flown is the sum of Leg1, Leg2, and Leg3:
Total distance flown = 55 miles (Leg1) + 25 miles (Leg2) + Unknown (Leg3)
The original flight plan's distance is only 55 miles.
Therefore, the extra miles flown by the pilot = Total distance flown - Original distance
= (55 + 25 + Unknown) miles - 55 miles
= 80 miles + Unknown - 55 miles
= 80 miles - 55 miles + Unknown
= 25 miles + Unknown
So, the pilot has flown at least 25 miles beyond his original flight plan. The exact number of extra miles depends on the distance of Leg3 to reach the original destination.
To solve this problem, we can utilize vector addition to determine the pilot's total displacement and the distance he needs to fly the next day.
Let's break down the flight path into different steps:
Step 1: Flying 55 miles west
The pilot flies 55 miles west, which can be represented as a vector pointing directly to the west. We'll call this vector "V1".
Step 2: Flying 25 miles at 15 degrees east of south
The pilot then flies 25 miles at an angle of 15 degrees east of south. This can be represented as a vector pointing in the direction 15 degrees east of south with a magnitude of 25 miles. We'll call this vector "V2".
Step 3: Finding the total displacement
To find the total displacement, we need to add the vectors V1 and V2. We can do this by breaking them down into their horizontal (west-east) and vertical (north-south) components and then adding them separately.
Let's calculate the horizontal and vertical components of V2:
Horizontal component of V2 = 25 * cos(15 degrees)
Vertical component of V2 = 25 * sin(15 degrees)
Now we can add the components:
Horizontal component of displacement = Horizontal component of V1 + Horizontal component of V2
Vertical component of displacement = Vertical component of V1 + Vertical component of V2
Step 4: Determine the distance and direction to fly the next day
To go directly to the original destination, the pilot's next flight must cancel out the displacement caused by flying to the third airport. This means flying in the opposite direction (west-east) and canceling out the vertical displacement (north-south). We can find the magnitude of this displacement using the Pythagorean theorem.
The distance to fly the next day = sqrt((horizontal component of displacement)^2 + (vertical component of displacement)^2)
The direction to fly the next day can be calculated using trigonometry. We can use the inverse tangent function to find the angle between the horizontal displacement and the vertical displacement.
Now let's substitute the values we calculated:
Distance to fly the next day = sqrt((Horizontal component of displacement)^2 + (Vertical component of displacement)^2)
Direction to fly the next day = atan(Vertical component of displacement / Horizontal component of displacement)
Finally, to calculate the extra miles flown beyond the original flight plan, we need to subtract the distance between the third airport and the original destination from the total distance flown:
Extra miles flown beyond the original flight plan = Distance flown - Distance between third airport and original destination
Now you have the steps to calculate the answers to both parts of the question.
15 deg. E of S = 285 deg CCW from 0 deg
X = HOR = 25cos(285) - 55 = -48.5,
Y = VER = 25sin(285) = -24.15,
tanA = Y/X = -24.15/-48.5 = 0.4979,
A = 26.47 deg. = 206.47 deg.(3rd quad.)
d = Y/sinA = -24.15/sin(206.5)=54.18mi
d = 54.18 mi @ 206.47 deg.
a. 54.18 mi N @ 63.5 deg W of South
Location: 3rd QUADRANT.
B. D = 2 * 54.18 = 108.36 MILES