A plane flies 30 mi on a bearing of 200 degrees and then turns and flies 40 mi on a bearing of 10 degrees. Find the resultant displacement vector as a distance and bearing.
resultant
= 30(cos200, sin200) + 40(cos10, sin10)
= (-28.19-10.26) + (39.39 , 6.946
= (11.20 , -3.315)
magnitude = √(11.20^2 + (-3.315)^2) = 11.68 miles
bearing:
tanØ = -3.315/11.2 = ...
Ø = 343.5°
Well, it seems like this plane is really good at going in circles! Let's see if we can figure out where it ended up.
To find the resultant displacement vector, we can treat the two legs of the plane's flight as two sides of a triangle. The third side of the triangle will be the displacement vector we're looking for.
First, let's use some trigonometry to determine the horizontal and vertical components of each leg of the flight.
For the first leg, the horizontal component can be found using the cosine of the angle, which is cos(200 degrees). Multiplying this by the length of the leg of the flight, 30 mi, we get the horizontal component to be -28.945 mi (rounded to three decimal places).
The vertical component of the first leg can be found using the sine of the angle, which is sin(200 degrees). Multiplying this by 30 mi, we get the vertical component to be -9.807 mi (rounded to three decimal places).
Now for the second leg, the horizontal component can be found using the cosine of the angle, which is cos(10 degrees). Multiplying this by the length of the leg, 40 mi, we get the horizontal component to be 39.836 mi (rounded to three decimal places).
The vertical component of the second leg can be found using the sine of the angle, which is sin(10 degrees). Multiplying this by 40 mi, we get the vertical component to be 6.950 mi (rounded to three decimal places).
Now we need to find the horizontal and vertical components of the displacement vector by adding up the components of the two legs:
Horizontal component = -28.945 mi + 39.836 mi = 10.891 mi (rounded to three decimal places)
Vertical component = -9.807 mi + 6.950 mi = -2.857 mi (rounded to three decimal places)
To find the distance of the displacement vector, we can use the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. So the distance can be found using:
Distance = √((horizontal component)^2 + (vertical component)^2)
= √((10.891 mi)^2 + (-2.857 mi)^2)
≈ √(118.617 mi^2 + 8.172 mi^2)
≈ √126.789 mi^2
≈ 11.271 mi (rounded to three decimal places)
So, the distance of the resultant displacement vector is approximately 11.271 mi.
Now let's find the bearing of the displacement vector. We can use some trigonometry again:
Bearing = arctan(vertical component / horizontal component)
Bearing = arctan((-2.857 mi) / (10.891 mi))
≈ arctan(-0.262)
≈ -14.475 degrees (rounded to three decimal places)
Since the bearing is negative, we can add 360 degrees to get a positive angle:
Bearing ≈ 360 degrees - 14.475 degrees
≈ 345.525 degrees (rounded to three decimal places)
So, the resultant displacement vector is approximately 11.271 mi on a bearing of 345.525 degrees.
But remember, this is all hypothetical and based on my calculations. In reality, the plane might just be circling around, having a grand old time without getting anywhere!
To find the resultant displacement vector, we need to find the sum of the individual displacement vectors.
First, let's find the x and y components of the first displacement vector.
The distance is 30 mi, and the bearing is 200 degrees.
To find the x component, we can use the cosine function:
x1 = 30 * cos(200°)
To find the y component, we can use the sine function:
y1 = 30 * sin(200°)
Using a calculator, we find:
x1 ≈ -25.674 mi
y1 ≈ 14.788 mi
Next, let's find the x and y components of the second displacement vector.
The distance is 40 mi, and the bearing is 10 degrees.
To find the x component:
x2 = 40 * cos(10°)
To find the y component:
y2 = 40 * sin(10°)
Using a calculator, we find:
x2 ≈ 39.848 mi
y2 ≈ 6.881 mi
To find the resultant x and y components, we can add the corresponding components together:
x = x1 + x2
y = y1 + y2
x ≈ -25.674 + 39.848 ≈ 14.174 mi
y ≈ 14.788 + 6.881 ≈ 21.669 mi
To find the resultant distance, we can use the Pythagorean theorem:
distance = sqrt(x^2 + y^2)
distance ≈ sqrt(14.174^2 + 21.669^2) ≈ sqrt(229.415 + 468.945) ≈ sqrt(698.36) ≈ 26.44 mi
To find the resultant bearing, we can use the inverse tangent function:
bearing = atan(y/x)
bearing ≈ atan(21.669/14.174) ≈ atan(1.526) ≈ 58.587°
Therefore, the resultant displacement vector is approximately 26.44 mi on a bearing of 58.587 degrees.
To find the resultant displacement vector, we need to calculate both the distance and the bearing.
1. Distance:
To find the distance, we can use the Pythagorean theorem. The plane flew a distance of 30 miles on a bearing of 200 degrees and then 40 miles on a bearing of 10 degrees. These form two sides of a right triangle.
Using the formula c^2 = a^2 + b^2, where c is the hypotenuse (resultant displacement), and a and b are the two sides, we can solve for c:
c^2 = 30^2 + 40^2
c^2 = 900 + 1600
c^2 = 2500
Taking the square root of both sides:
c = √2500
c = 50
Therefore, the resultant displacement (distance) is 50 miles.
2. Bearing:
To find the bearing, we need to determine the angle between the displacement vector and the reference axis (north).
First, let's find the angle formed by the two sides of the right triangle we previously calculated. This angle will help us determine the bearing.
Using the law of cosines, we can solve for the angle:
c^2 = a^2 + b^2 - 2ab * cos(C)
Where C is the angle opposite side c:
50^2 = 30^2 + 40^2 - 2 * 30 * 40 * cos(C)
2500 = 900 + 1600 - 2400 * cos(C)
2500 = 2500 - 2400 * cos(C)
To solve for cos(C):
cos(C) = 2500 / 2400
C ≈ cos^(-1)(1.0417)
C ≈ 43.6 degrees
To determine the bearing, we need to convert this angle into a bearing (measured clockwise from north).
Since the plane initially started on a bearing of 200 degrees, we need to add this to the angle we calculated:
Bearing = 200 + 43.6
Bearing ≈ 243.6 degrees
Therefore, the resultant displacement (bearing) is approximately 243.6 degrees.