Asked by Joann
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.
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.
Answers
Answered by
Reiny
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.
= (-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.
Answered by
henry2,
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.
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.
Answered by
henry2,
Correction: -73o = 73o E. of S. = 107o CW.
Answered by
Joann
How was the angle 45o calculated?
Answered by
henry2,
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.
The wind was headed north-east which is 45o E. of N. or 45o CW
from +y-axis.
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