A plane flies at 250 km/h [N]. After flying for 2.0 h the plane has been blown off course 40 km [SW]. What is the resulting velocity of the plane?
Initailly the velocity is 250Km/h to the North.
After blown off apart from the 250km/h to the North we have a velocity component of 40Km/h to south west.
So if we convert the SW velocity to components South and west we will get;
Velocity component to west = 40*cos45
Velocity component to south = 40*sin45 ;same thing can be expressed as -40*sin45 to the north.
Now we can add up Velocity component to North.
Then resulting velocity to North = 250+(-40*sin45) km/h
= 221.716 km/h
Now we have two velocities;
first one to the North(221.716) and next to the west(40*sin45).
So the resultant velosity = sqrt(221.716+40*sin45)
The direction of resultant force= tan-1(221.716/40*sin45)
= 82.73 degrees from west
Therefore the final velocity is 223.513 km/h to a direction 82.73 degrees from west.
- At the blown off still the plane remain its 250 km/h [N]
As I understood the problem, it is the distance of 40 Km SW to which the plane is blown off course from the point it would have reached after 2 hours i.e. 2x250 = 500 km due North.
In this situation the blown off position to from 500km N is
towards South = 40.cos(45) = 28.28km
towards West = 40.sin(45) = 28.28km
Resultant position = 500-28.28 = 471.72 N and 28.28 W
Resultant direction = tan-1(28.28/471.72) = 3.43 degree NW
Distance covered in 2 hours = sqrt(471.72^2+28.28^2) = 472.567km
Resultant velocity = 472.567/2 = 236.28km/h
The resultant velocity of the plane is 236.28 km/h in direction 3.43 degrees NW