Vectors applied math

Question

You swim at 3km/h directly across a stream with a current of 5km/h. Your actual velocity is the vector sum of the streams velocity snd your swimming velocity. Find the magnitude of your actual velocity .

 

 

2) An aircraft is flying at 300km/h on a bearing of 80 degrees. A wing is blowing at 50km/h from a bearing of 120 degrees . Determine the actual ground speed and direction of the aircraft. 

Answer 1

For the swimmer problem:

 

Let's arbitrarily assume the stream is moving in a +X DIR. 

the swimmer moving directly across, would be moving perpendicular

to the stream current in X DIR,   or +Y DIR

 

i'll try to diagram it below

 

--------------------->  current 5 mph (+X)

^

|

| swimmer 3 mph (+Y)

|

 

the swimmer would then also be moving in the current at 5 mph so we add the vectors

Graphically, you add the vector tail to the head

 

The resultant is a somewhat diagonal direction, or hypotenuse of the two vectors (sides of right triangle)

 

the resultant vector has a magnitude of SQRT(3² + 5²) or

the SQRT(34)  OR 5.83 mph .  

 

We could also find the angle from +Y as  ArcTAn(5/3) or 59.0 deg from vertical (or +Y)

 

====================================================

 

For the airplane has a bearing of 80 deg at 300 mph, with a wind bearing of 120 deg at 50 mph. 

I wont try to diagram.  

 

 

The steps are:

1) Break each vector into it's X and Y components. 

2) add the X and Y separately. These then make up the sides of a right triangle.

3) the resultant vector is found using Pythagorean Theorem

4) the resultant bearing vector is ArcTan(X/Y). this is the angle from north (bearing)

 

since the bearing is from north (+y)  the Y comp is COSINE, X component is SINE  

 

so, 1)  airplane   X Vector =  300*SIN(80)   Y Vector = 300*COS(80)

          wind        X Vector =   50*SIN(120) Y Vector =   50*COS(120)

 

      OR  Airplane X = 295.44   Y =  52.09

            Wind      X =  43.30    Y = -25.00

 

    2) SUM =    X = 338.74   Y = 27.09

 

    3) the resultant vector magnitude = SQRT(X² + Y²)

        VECMAG = 339.82 MPH

 

    4) the bearing would be ArcTan(X/Y)  or  85.43 DEG FROM NORTH. 

 

 

*** NOTE: for math, the ArcTAN fctn normally has the argument Y/X which

is an angle from the X axis.  However for BEARING, it is the angle from north,

or the +Y axis, therefore the ArcTAn argument is X/Y