- AN AIRPLANE IS HEADING DUE NORTH AT AN AIR
SPEED OF 700 KILOMETERS PER HOUR,
BUT THERE IS A WIND BLOWING FROM THE NORTHEAST
AT 60 KILOMETERS PER HOUR.
HOW MANY DEGREES OFF COURSE WILL THE PLANE FLY
AND WHAT IS THE PLANE'S SPEED RELATIVE TO THE GROUND?
LET'S START BY MODELING THE PLANE AND THE WIND
USING TWO VECTORS.
SO THE BLUE VECTOR REPRESENTS THE SPEED
AND DIRECTION OF THE PLANE.
AND SINCE IT'S GOING 700 KILOMETERS PER HOUR DUE NORTH,
THE MAGNITUDE OF THIS VECTOR IS 700.
THE WIND IS BLOWING OUT OF THE NORTHEAST,
SO IT'S BLOWING IN THE DIRECTION OF SOUTHWEST.
SO THE MAGNITUDE OF THE RED VECTOR WOULD BE 60.
SO OUR ULTIMATE GOAL HERE IS TO ADD THESE TWO VECTORS TOGETHER,
THEN DETERMINE THE DIRECTION AND MAGNITUDE OF THAT VECTOR.
TO ADD THESE TWO VECTORS WE WANT THEM BOTH IN COMPONENT FORM.
WELL, FOR VECTOR P, SINCE IT'S HEADING DUE NORTH,
WE KNOW THE HORIZONTAL COMPONENT WOULD BE ZERO,
AND THE VERTICAL COMPONENT WOULD BE 700.
BUT FOR VECTOR W, WE'LL HAVE TO USE THESE TWO FORMULAS HERE
TO DETERMINE THE HORIZONTAL AND VERTICAL COMPONENT.
SO WE NEED TO FIRST RECOGNIZE THAT OUR ANGLE THETA,
MEASURED FROM THE .+X AXIS WOULD BE 225 DEGREES,
SINCE IT'S BLOWING OUT OF THE NORTHEAST.
AND THEN SINCE THE MAGNITUDE IS 60,
WE CAN FIND BOTH THE HORIZONTAL AND VERTICAL COMPONENTS.
SO FOR VECTOR W, X = 60 x COSINE 225 DEGREES,
AND Y = 60 x SINE 225 DEGREES.
BECAUSE WE'RE IN THE THIRD QUADRANT,
WHERE BOTH X AND Y ARE NEGATIVE,
BOTH THE COSINE AND SINE FUNCTIONS VALUE
WILL BE NEGATIVE.
WE SHOULD ALSO RECOGNIZE
THAT WE'RE GOING TO HAVE A 45 DEGREE REFERENCE ANGLE,
OR IF WE LOOK AT THE UNIT CIRCLE,
NOTICE BOTH COSINE 225 DEGREES AND SINE 225 DEGREES
IS -SQUARE ROOT 2/2.
SO WE HAVE X = 60 x -SQUARE ROOT 2/2, AND THE SAME THING FOR Y.
SO BOTH OF THESE SIMPLIFY TO (-30, SQUARE ROOT 2).
SO VECTOR W, IN COMPONENT FORM,
WOULD BE (-30 SQUARE ROOT 2, -30 SQUARE ROOT 2).
SO NOW WE CAN ADD VECTOR P AND VECTOR W.
THE HORIZONTAL COMPONENT WILL BE 0
+ (-30, SQUARE ROOT 2), OR JUST (-30, SQUARE ROOT 2).
AND THE VERTICAL COMPONENT WILL BE 700 + (-30, SQUARE ROOT 2),
OR 700 - (30, SQUARE ROOT 2).
NOW THAT WE HAVE THE SUM OF THESE TWO VECTORS,
THE MAGNITUDE OF THIS VECTOR WILL GIVE US THE SPEED RELATIVE
TO THE GROUND.
AND THEN IF WE DETERMINE THETA, THAT WILL GIVE US THE DIRECTION,
WHICH WE CAN THEN DETERMINE
HOW FAR OFF COURSE THE PLANE IS FLYING.
SO NOW WE'LL USE THESE TWO FORMULAS TO ANSWER OUR QUESTION.
LET'S START WITH FINDING OUR DIRECTION ANGLE THETA.
SO TANGENT THETA = TO Y DIVIDED BY X
OR 700 - (30, SQUARE ROOT 2) DIVIDED BY (-30, SQUARE ROOT 2).
BUT BEFORE WE DO THIS IT'S IMPORTANT TO RECOGNIZE
THAT IF WE WERE TO ADD THESE TWO VECTORS TOGETHER,
NOTICE HOW THE RESULTANT VECTOR, OR THE SUM OF THE TWO VECTORS,
WOULD BE IN THE SECOND QUADRANT,
THEREFORE, THETA MUST ALSO BE IN THE SECOND QUADRANT.
LET'S MAKE SURE WE'RE IN DEGREE MODE, WHICH WE ARE.
SO WE'LL PRESS 2nd, TANGENT,
AND THEN THE NUMERATOR IS GOING TO BE 700 - (30, SQUARE ROOT 2).
THERE'S OUR NUMERATOR, DIVIDED BY (-30, SQUARE ROOT 2).
SO NOTICE HOW THETA IS APPROXIMATELY -86.31 DEGREES,
BUT THIS IS ACTUALLY IN THE FOURTH QUADRANT.
BUT WE CAN USE THIS TO FIND OUR ANGLE IN THE SECOND QUADRANT.
AGAIN, THE CALCULATOR GAVE US THIS ANGLE HERE
IN THE FOURTH QUADRANT.
BUT SINCE WE KNOW THE TANGENT FUNCTION VALUE
IS ALSO NEGATIVE IN THE SECOND QUADRANT,
AND THAT'S WHERE OUR ANGLE SHOULD BE,
IF WE SKETCH AN 86.31 DEGREE REFERENCE ANGLE
IN THE SECOND QUADRANT,
WE CAN FIND THE ANGLE WE NEED IN THE SECOND QUADRANT.
SO THIS IS THE TERMINAL SIDE OF OUR ANGLE.
THEN THE REFERENCE ANGLE HERE MUST ALSO BE 86.31 DEGREES.
SO WE CAN FIND OUR ANGLE THETA BY TAKING HALF A ROTATION,
OR 180 DEGREES,
AND THEN SUBTRACTING THE REFERENCE ANGLE
OF 86.31 DEGREES, WHICH IS 93.69 DEGREES.
NOW, IF I GO BACK TO THE QUESTION FOR A MOMENT,
THEY'RE ASKING US,
HOW MANY DEGREES OFF COURSE WILL THE PLANE FLY?
SO OUR ANSWER IS NOT GOING TO BE 93.69 DEGREES.
THE PLANE IS SUPPOSED TO BE FLYING DUE NORTH
AT A DIRECTION OF 90 DEGREES.
AND SO THE DIFFERENCE BETWEEN 93.69 AND 90
IS HOW FAR OFF COURSE THE PLANE WILL FLY.
SO THE PLANE WILL FLY 3.69 DEGREES OFF COURSE,
WITH THE WIND BLOWING OUT OF THE NORTHEAST AT 60 MILES PER HOUR.
NOW, FOR THE SECOND QUESTION,
WE WANT TO FIND THE PLANE'S SPEED RELATIVE TO THE GROUND,
WHICH WOULD BE THE MAGNITUDE OF OUR RESULTANT VECTOR HERE.
LET'S GO AHEAD AND FIND THE MAGNITUDE OF THIS
ON THE NEXT SLIDE.
AND SO THE MAGNITUDE
WILL BE EQUAL TO THE SQUARE ROOT OF X SQUARED + Y SQUARED,
OR (-30, SQUARE ROOT 2 SQUARED)
+ 700 - (30, SQUARE ROOT 2 SQUARED).
AND NOW WE'LL GO BACK TO THE CALCULATOR.
SO THE MAGNITUDE IS APPROXIMATELY 658.94,
WHICH MEANS THE PLANE'S SPEED RELATIVE TO THE GROUND
WOULD BE APPROXIMATELY 658.94 KILOMETERS PER HOUR.
SO IF WE GO BACK AND TAKE A LOOK AT THE VECTORS
JUST FOR A MOMENT,
BECAUSE OF THE DIRECTION OF THE RED VECTOR,
WE CAN SEE THAT IT SHOULD SLOW THE PLANE DOWN,
AND THAT'S WHY THE RELATIVE SPEED
IS LESS THAN 700 MILES PER HOUR.
AND ALSO, THE WIND BLOWING IN THIS DIRECTION
WOULD MAKE THE PLANE FLY SLIGHTLY WEST OF DUE NORTH,
AND THAT'S WHY THE PLANE IS FLYING
APPROXIMATELY 3.69 DEGREES OFF COURSE INTO QUADRANT TWO.
OKAY, I HOPE THIS EXPLANATION WAS HELPFUL.