An airline route from San Francisco to Honolulu is on a bearing of 233.0°. A jet flying at 450 mph on that bearing encounters a wind blowing at 39.0 mph from a direction of 114.0°. Find the resulting bearing and ground speed of the plane.
To solve this problem, we can use vector addition.
Step 1: Convert the angles to vectors.
The jet's velocity vector is given by v_jet = 450 mph at 233.0°.
The wind's velocity vector is given by v_wind = 39.0 mph at 114.0°.
Step 2: Break down the vectors into their components.
For the jet's velocity vector, the x-component is v_jet_x = 450 * cos(233.0°) and the y-component is v_jet_y = 450 * sin(233.0°).
For the wind's velocity vector, the x-component is v_wind_x = 39.0 * cos(114.0°) and the y-component is v_wind_y = 39.0 * sin(114.0°).
Step 3: Add the vector components.
To find the resulting velocity vector, add the x-components and the y-components separately.
v_result_x = v_jet_x + v_wind_x
v_result_y = v_jet_y + v_wind_y
Step 4: Find the magnitude and direction of the resulting velocity vector.
The magnitude (ground speed) of the resulting velocity vector is given by the Pythagorean theorem:
ground speed = sqrt(v_result_x^2 + v_result_y^2)
The direction (bearing) of the resulting velocity vector is given by:
bearing = arctan(v_result_y / v_result_x)
Step 5: Calculate the values.
Using a calculator or computer program, evaluate the equations:
v_jet_x = 450 * cos(233.0°) = -315.522 mph
v_jet_y = 450 * sin(233.0°) = 340.253 mph
v_wind_x = 39.0 * cos(114.0°) = -11.751 mph
v_wind_y = 39.0 * sin(114.0°) = 29.783 mph
v_result_x = v_jet_x + v_wind_x = -315.522 mph - 11.751 mph = -327.273 mph
v_result_y = v_jet_y + v_wind_y = 340.253 mph + 29.783 mph = 370.036 mph
ground speed = sqrt(v_result_x^2 + v_result_y^2) = sqrt((-327.273 mph)^2 + (370.036 mph)^2) ≈ 487 mph
bearing = arctan(v_result_y / v_result_x) = arctan(370.036 mph / -327.273 mph) ≈ 130.8°
Step 6: Round the answers.
The resulting bearing is approximately 130.8°.
The ground speed of the plane is approximately 487 mph.
Therefore, the resulting bearing is 130.8° and the ground speed of the plane is 487 mph.