n = speed north
n - 100 = speed east
n^2 + (n - 100)^2 = 500^2
2 n^2 - 200n + 10000 - 250000 = 0
n^2 - 100 n - 120000 = 0
solve for n
the north distance is 400√2
time = distance / speed
the bearing (Θ , east of north)
... tan(Θ) = (n - 100) / n
2 Dimensional Motion-Kinematics
QS 1: An airplane wishes to travel 800 km northeast of a starting point. There is a wind blow towards the east at 100km/h. Assuming the maximum speed of the plane in still air is 500km/h,
a: what course should the pilot steer to arrive in minimum time?
b: How long does the trip take?
2 answers
a. Vp + 100 = 500[45o].
Vp + 100 = 500*sin45 + i500*Cos45,
Vp + 100 = 353.6 + 353.6i,
Vp = 253.6 + 353.6i = 435km/h[35.6o]E. of N.
Tan A = X/Y.
b. d = V*T.
800 = 435 * T,
=
Vp + 100 = 500*sin45 + i500*Cos45,
Vp + 100 = 353.6 + 353.6i,
Vp = 253.6 + 353.6i = 435km/h[35.6o]E. of N.
Tan A = X/Y.
b. d = V*T.
800 = 435 * T,
=
1 answer