vector and velocity?

Question

2 airplanes, A & B are flying at the same altitude in the direction N30˚E and N10˚W. when B is 10km to the east of A, the pilots of both airplanes found that their planes are in the path of collision. If the speed of A is 500km h-1, find the speed of B. Give your answer to 3 significant figures. If at that point in time, B changes direction to N45˚W without changing its speed, find the closest distance between A & B.
(ANS:440 km h-1, 2.13 km)

Answer 1

The vector of velocity of a is
sin(30)*500,
cos(30)*500

The vector velocity of B is
sin(10)*va,
cos(10)*va

Since B is east of A by 10 km, they must travel the same distance North to collide, so
cos(30)*500/cos(10)=va
or
440 km/hr

The distance of separation after the B course change is the square root sum of squares of the velocity vectors. Setting North as positive y and East as positive x,
delta x=t*(sin(30)*500+sin(45)*440)-10
and
delta y=t*(cos(30*500)-cos(45)*440)
using t=0 as the moment that B changes course which is when B is 10km due East of A.
the distance then is
10*sqrt(3300*t^2-112*t+1)
taking d(distance)/dt using the chain rule and setting the first derivative equal to zero:
6600*t-112=0
or
t=0.017 hr
plug that into the separation equation to get
d=2.13 km

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