A plane is flying with an airspeed of 190 miles per hour and heading 140°. The wind currents are running at 25 miles per hour at 165° clockwise from due north. Use vectors to find the true course and ground speed of the plane. (Round your answers to the nearest ten for the speed and to the nearest whole number for the angle.)
ground speed = (190cos140°, 190sin140°) + (25cos165°,25sin165°)
= (-169.7, 128.6)
magnitude = √( (-169.7)^2 + (128.6^2) ) = appr 212.9 mph
for direction:
tanØ = -128.6/169.7
angle Ø in standard position = 37.16°
or
after sketching the parallelogram and labelling the angles
|R|^2 = 190^2 + 25^2 - 2(25)(190)cos155°
= 45334.92..
R = √45334.92 = appr 212.9 km/h , just as above
The acute angle of the needed triangle, by the sine law:
sin x /25 = sin155/212.9
angle x = 2.84°
add that to 140° to get a heading of 142.84°
Notice from my vector calculation : 180-37.16 ° = 142.84°
just add the vectors. I get
212.92 @ 142.84°
To find the true course and ground speed of the plane, we first need to break down the airspeed and wind speed into their respective components.
Let the x-axis represent eastward direction and the y-axis represent northward direction.
For the airspeed:
- The eastward component (x) is given by 190 * cos(140°).
- The northward component (y) is given by 190 * sin(140°).
For the wind speed:
- The eastward component (x) is given by 25 * cos(165°).
- The northward component (y) is given by 25 * sin(165°).
Next, we can add the components of the airspeed and wind speed to find the resultant vector, which represents the actual velocity of the plane.
- The eastward component of the resultant vector is the sum of the x-components of airspeed and wind speed.
- The northward component of the resultant vector is the sum of the y-components of airspeed and wind speed.
Finally, we can use the Pythagorean theorem to find the magnitude of the resultant vector, which represents the ground speed of the plane. The angle of the resultant vector with respect to the x-axis represents the true course of the plane.
Calculating the components:
Airspeed:
- x-component: 190 * cos(140°) ≈ -65.00
- y-component: 190 * sin(140°) ≈ 161.77
Wind speed:
- x-component: 25 * cos(165°) ≈ 9.47
- y-component: 25 * sin(165°) ≈ -23.84
Resultant vector (velocity):
- x-component: -65.00 + 9.47 ≈ -55.53
- y-component: 161.77 - 23.84 ≈ 137.93
Calculating the ground speed and true course:
Magnitude (ground speed) = √((-55.53)^2 + (137.93)^2) ≈ 149.00 miles per hour
Angle (true course) = arctan(137.93 / -55.53) ≈ -68° (nearest whole number)
Therefore, the true course of the plane is approximately 68° (clockwise from due north) and the ground speed is approximately 149 miles per hour.
To find the true course and ground speed of the plane, we need to consider the effect of wind on the plane's motion. We can use vector addition to find the resultant velocity.
Let's break down the given information:
The airspeed of the plane is 190 miles per hour, and it is heading 140°. This gives us a vector representing the velocity of the plane in still air. We'll call it A.
The wind currents are running at 25 miles per hour, and the direction is 165° clockwise from due north. This gives us a vector representing the velocity of the wind. We'll call it W.
To find the true velocity of the plane, we need to add the velocity of the plane in still air (A) to the velocity of the wind (W). Setting up coordinate axes, we can resolve the vectors into their components.
For the plane's velocity in still air, we can use trigonometry to find its x and y components. Since the angle is measured clockwise from due north, we subtract it from 90° to get the angle measured counterclockwise from the positive x-axis.
The x-component of the plane's velocity is:
Vx = 190 * cos(90° - 140°) ≈ -141.42 mph
The y-component of the plane's velocity is:
Vy = 190 * sin(90° - 140°) ≈ 131.38 mph
For the wind's velocity, we also need to use trigonometry. The angle is measured clockwise from due north, so we add 90° to it to get the angle measured counterclockwise from the positive x-axis.
The x-component of the wind's velocity is:
Wx = 25 * cos(90° + 165°) ≈ -23.91 mph
The y-component of the wind's velocity is:
Wy = 25 * sin(90° + 165°) ≈ 20.24 mph
The resultant velocity, which is the sum of the velocities of the plane and wind, can be found by adding the respective components.
The x-component of the resultant velocity is:
Rx = Vx + Wx ≈ -141.42 mph -23.91 mph = -165.33 mph
The y-component of the resultant velocity is:
Ry = Vy + Wy ≈ 131.38 mph + 20.24 mph = 151.62 mph
To find the magnitude of the resultant velocity (ground speed), we can use the Pythagorean theorem:
|R| = sqrt(Rx^2 + Ry^2) ≈ sqrt((-165.33)^2 + (151.62)^2) ≈ 220.0 mph
The true course (direction of the ground speed) can be found using the inverse tangent function:
θ = tan^(-1)(Ry/Rx) ≈ tan^(-1)(151.62 mph/-165.33 mph) ≈ -43.9°
Since the question asks for the angle in a clockwise direction from due north, we'll convert the angle to a clockwise measurement:
True course (angle) = 360° - (-43.9°) ≈ 404°
Therefore, the true course of the plane is approximately 404°, and the ground speed is approximately 220 mph.