Physics Problem, Grade 12?
An airplane flies at an airspeed of 441 km/hr. The pilot wants to fly 799 km to the north. She knows that she must head 19.6° east of north to fly directly there. If the plane arrives in 2.34 hr, find the magnitude of the wind velocity vector.
Can't Seem to find anyone that can solve this problem,
Can you guys help out?
Comments
The resultant speed of the trip is given by...
799 km / 2.34 h = 341.45 km/h
This is the magnitude of the total vector, the angle is 19.6°
This vector is the sum of the vecotr of the airplane plus the vector of the wind.
Va = (441 km/h) j^
Vw = (Vwx) i^ + (Vwy) j^
Vr = (341.45sin(19.6°)) i^ + (341.45cos(19.6°)) j^
Adding each component...
441 + Vwy = 341.45cos(19.6°)
Vwy = -119 km/h
Vwx = 341.45sin(19.6°)
Vwx = 115 km/h
The wind velocity vector is...
Vw = 115 i^ - 119 j^
The magnitude is √( (115)² + (-119)² ) = 165 km/h <===============
799 km N / 2.34 h = 341.45 km/h N
this is the resultant
draw this vector (up on y axis) to scale then draw 441 at 19.6 deg right of y axis
connect the ends of these two vectors
make the resultant the diagonal of a parallelogram by drawing the parallels to your two already drawn sides ( [email protected] and the tip connector.
The side which points into quadrant three - a bit south of west - is the wind speed
Solve for the magnitude and angle of this vector
this is a vector subtraction
341.45@ 0 deg - 441@ 19.6 deg = 165.4 @ -116.6 deg or 165 km/h, 26.6 deg S of W
that's a pretty substantial headwind - real hard on the old fuel milage
Pulse AB will retrace its path