Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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"?