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A particle possesses two velocities at the same time one 4 m/s due south and other the 2√2 m/s due north- east. Find the magnitude and direction of resultant velocity.
Velocity components 4 m/s and 2√2 m/s acting on the object are shown in figure.

Resultant velocity is detrmined by parallelogram method .

Let velocity components OA = 4 m/s and OB = 2√2 m/s are two adjacent sides of parallelogram as shown in figure.

Let us make the parallelogram OACB with adjacent sides OA and OB as shown in figure.

Diagonal OC is resultant velocity that is determined using cosine formula from ΔOBC

OC = OB2 + BC2 - (2 × OB × BC × cos45)

OC2 =  8 + 16 - ( 2 × 4 × 2√2 × cos45 ) = 8

Hence , OC = √8 m/s  = 2√2 m/s

Since OB = OC , we get OBC = OCB = AOC = 45o
Hence magnitude of resultant = 2√2 m/s and direction is South-East

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Resultant OC is determined by vector method as explained below

=  - 4  m/s
=  ( 2   + 2  )  m/s
and  are unit vectors along x-axis and y-axis direction
Resultant  =  +   = ( 2   - 2  )  m/s
Magnitude of resultant = |  | = ( 22 + 22 )1/2 m/s =  2√2 m/s .
Direction of resultant = tan-1 ( uy/ ux ) = tan-1 ( -2/ 2 ) = -45o ,

Where uy is y-component of resultant and ux is x-component of resultant

i.e. resultant velocity makes angle 45o with x-axis in clockwise direction .

Hence direction of resutlat is South-East
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