A = P + W
P = A - W
............./|
..... W /..|
......./.....|
.../45°...|
C---------|-------- E
.||........|
.|.|......| P
.|A|....|
.|...|..|
.|....||
.|....N
S
Since the plane will fly for (6:30 - 4:00) = 2.5 hr in a 40 km/hr wind (I assume you meant 40 km/hr, not 40 km), the length of the wind vector is 40 km/hr * 2.5 hr = 100 km, and the direction is to the northeast. So, the north component of the wind vector W is
100 km * sin 45° = 100 km * √2 / 2 = 50 km * √2
The east component of the wind vector W is
100 km * cos 45° = 100 km * √2 / 2 = 50 km * √2
Remember, north is the opposite of south, so we can write the wind vector as
W = -50 km * √2 S + 50 km * √2 E
So, the vector the plane should fly at is
P = A - W
P = (225 km * √2 - 270.9 km) * E + 225 km * √2 S - (-50 km * √2 S + 50 km * √2 E)
P = (225 km * √2 - 270.9 km - 50 km * √2) * E + (225 km * √2 + 50 km * √2) * S
P = (175 km * √2 - 270.9 km) * E + (275 km * √2) * S
P = (247.5 km - 270.9 km) * E + 388.9 km * S
P = -23.4 km * E + 388.9 km * S
That's the vector. We need to convert that to a speed and a direction. Direction is easy; we can find the angle a below from trig.
...............|
........23.4.|
........-------|-------- E
........|..../.|
........|.../a|
388.9|../...|
........|./....|
........|/.....|
..............S
tan a = 23.4 km / 388.9 km
a = atan(23.4 / 388.9)
a = 3.4°
So, we need to travel 3.4° west of south, or at a compass heading of 183.4°.
Now, what about the speed? Speed is simply distance divided by time, and we know the time is 2.5 hr. All we need is the distance, which is the magnitude of vector P. We'll use the Pythagorean theorem.
d = √[(-23.4 km)^2 + (388.9 km)^2]
d = √[547.56 km^2 + 151243.21 km^2]
d = √[151790.77 km^2]
d = 389.6 km
So, the airspeed is
d / t = 389.6 km / 2.5 hr = 155.8 km/hr
I hope that helps!