Two airplanes leave an airport at the same
time. The velocity of the first airplane is
650 m/h at a heading of 40◦
. The velocity of
the second is 560 m/h at a heading of 108◦
.
How far apart are they after 1.5 h?
Answer in units of m.
Pleaase heelp
d1 = V1*T = 650[40o] * 1.5 = 975 m.[40o].
d2 = V2*T = 560[108o] * 1.5 = 840 m[108o].
d = d1-d2,
d = 975[40o] - 840[108o].
X = 975*Cos40 - 840*Cos108 =
Y = 975*sin40 - 840*sin108 =
d = sqrt(x^2 + y^2).