Velocity Triangles and Relative Velocity A typical problem: A helicopter can fly at 200 km h  1 in still air. It has to travel from an airport to a hospital. The hospital is 600 km from the airport on a bearing of 250 . The wind speed is 80 km h  1 and it is blowing from the south west. Work out the course that the helicopter must fly on and the time that it takes to get to the hospital.

First step: construct the velocity triangle  the resultant velocity must be in the direction of the hospital from the airport, i.e. the path that the helicopter has to fly along. Two things to look out for:  the direction of the vectors in the velocity triangle must follow this rule: resultant velocity  initial velocity  change in velocity e.g. Speed in still water or still air. This will be in the direction of the course that has to be set. This will usually be either the wind’s velocity or the current’s velocity.

Airport North Hospital This is the resultant velocity 250  North 45  Wind blowing FROM south west Helicopter’s speed in still air, in the direction of the course it must fly on.

110  70  45  135  155  Find the size of an internal angle of the triangle, using basic geometry - the north lines [dashed] are parallel.  Use Sine Rule to find  and hence  : sin   sin 155     9  7  [1 d.p.] and   15  3  Use Sine Rule again to find unknown side length [resultant speed] : 80  speed sin  sin   speed  125 km h  1 [3 s.f.]  Hence the helicopter must fly on a bearing of 240  [to nearest degree] and the time taken is 4 hours 48 minutes.