Asked by Alexa

An airplane is heading at a bearing of N23°E at a speed of 700 km/hr, but there is a wind blowing from the west at 60 km/hr. If the plane does not make any correction, what will be the bearing of the plane? What is its speed relative to the ground (ground speed = airspeed + wind speed). In order for the plane to fly to its original bearing, what correction needs to be made?

Answers

Answered by Steve
craft travel on a heading, not a bearing.

700 @ N23°E = (273.51,644.35)
60 @ E = (60,0)
Add them up and you have
(333.51,644.35) = 725.55 @ N27.36°E
Answered by Alexa
Can you explain how to get these from here:
700 @ N23°E = (273.51,644.35)

(333.51,644.35) = 725.55 @ N27.36°E
Answered by Damon
700 sin 23 = 273.51 = x speed (east)
700 cos 23 = 644.35 = y speed (north)

If x = 333.51
and y = 644.35

then resultant = sqrt (x^2+y^2)
and for angle east of north
tan angle = x/y
Answered by Henry
Vpw = Vp + Vw = 700[67o] + 60[0o] =
700*Cos67 + 700*sin67 + 60 =
273.5 + 644.4i + 60 = 333.5 + 644.4i =
725.6km/h[62.64o].

Vpw = Vp + Vw = 700[67o].
Vp + 60 = 700[67].
Vp = 273.5 + 644.4i - 60=213.5 + 644.4i
= 679km/h[71.7o] N. of E. = 18.3o E. of
N.

Answered by Ryan
Hi Henry, can you explain the second part of your answer please?
Is that the answer to the "What correction needs to be made" part?
What is that "i"?
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