A particular small drone has a speed in still air of 20ms-1. It is pointed in
the direction of the bearing 125 degrees, but there is a wind blowing at a speed of 7ms-1 from the south-west.
Take unit vectors i to point east and j to point north.
(a) Express the velocity d of the drone relative to the air and the velocity w
of the wind in component form, giving numerical values in ms-1 to two
decimal places.
(b) Express the resultant velocity v of the drone in component form, giving
numerical values in ms-1 to two decimal places.
(c) Hence find the magnitude and direction of the resultant velocity v of
the drone, giving the magnitude in ms-1 to two decimal places and the
direction as a bearing to the nearest degree.
v = 20(cos125,sin125) + 7(cos45,sin45)
= (-11.4715..., 16.38304..) + (4.9497..., 4.9497...)
= (-6.52178..., 21.33278...)
magnitude =√(42.5336... + 455.087..) = appr 22.307
tanØ = 21.332.../-6.52178... = -3.271...
since Ø must be between 90° and 180‚ Ø = appr 107°
the individual answers to your questions are contained in this solution.
All angles are measured CW from +y-axis.
a. Vd = 20m/s[125o] = 20*sin125 + (20*cos125)I = 16.38 - 11.47i.
Vw = 7m/s[45o] = 7*sin45 + (7*cos45)I = 4.95 + 4.95i.
b. Vr = 16.38-11.47i + 4.95+4.95i = 21.33 - 6.52i.
c. Vr = 21.33-6.52i = 22.30m/s[-73o].
-73o = 73o W. of N. = 287o CW.
Correction: -73o = 73o E. of S. = 107o CW.
How was the angle 45o calculated?
The wind was coming from south-west which is 45o w. of S.
The wind was headed north-east which is 45o E. of N. or 45o CW
from +y-axis.
(a) To express the velocity of the drone relative to the air (d) and the velocity of the wind (w) in component form, we can break them down into their east-west (x) and north-south (y) components.
Velocity of the drone relative to the air (d):
Given that the speed of the drone in still air is 20 m/s, and it is pointed in the direction of bearing 125 degrees, we can calculate the components as follows:
The east-west component (dx) can be calculated using trigonometry:
dx = speed x cos(angle)
dx = 20 x cos(125°)
dx = 20 x (-0.5736)
dx ≈ -11.47 m/s
The north-south component (dy) can also be calculated using trigonometry:
dy = speed x sin(angle)
dy = 20 x sin(125°)
dy = 20 x (0.8192)
dy ≈ 16.38 m/s
So the velocity of the drone relative to the air (d) is approximately d = -11.47i + 16.38j m/s.
Velocity of the wind (w):
Given that the wind is blowing at a speed of 7 m/s from the south-west, we can calculate its components as follows:
The east-west component (wx) would be the negative of the wind speed:
wx = -7 m/s
The north-south component (wy) would also be the negative of the wind speed, but we need to account for the wind direction being from the south-west. We can break it down into its south (negative) and west components:
wy = wind speed x (-sin(45°))
wy = 7 x (-0.7071)
wy ≈ -4.95 m/s
So the velocity of the wind (w) is approximately w = -7i - 4.95j m/s.
(b) To find the resultant velocity (v) of the drone, we need to add the velocity of the drone relative to the air (d) to the velocity of the wind (w):
v = d + w
v = (-11.47i + 16.38j) + (-7i - 4.95j)
v = -11.47i + (-7i) + 16.38j + (-4.95j)
v = -18.47i + 11.43j
So the resultant velocity of the drone (v) is v = -18.47i + 11.43j m/s.
(c) To find the magnitude and direction of the resultant velocity (v) of the drone, we can use the Pythagorean theorem to find the magnitude and trigonometry to find the direction.
Magnitude of v:
Magnitude = sqrt(i-component^2 + j-component^2)
Magnitude = sqrt((-18.47)^2 + 11.43^2)
Magnitude ≈ 21.83 m/s
Direction of v:
Direction = arctan(j-component / i-component)
Direction = arctan(11.43 / -18.47)
Direction ≈ -30.19 degrees
Therefore, the magnitude of the resultant velocity (v) of the drone is approximately 21.83 m/s, and the direction is a bearing of -30 degrees to the nearest degree.