Aboy runs 3.00 blocks north, 4.00 blocks northeast, and
5.00 blocks west. determine the length and direction of
the displacement vector that goes from the starting
point to his final position.

For this question, we determine first the x and y components of the distances. 
            3 blocks north:  
                             x - component = +3  ; y-component = 0
            4 blocks northeast:
                             x - component = (cos 45°)(4 blocks) = 2.83 blocks
                             y - component = (cos 45°)(4 blocks) = 2.83 blocks
           5 blocks west:
                              x-component = -5   ; y-component = 0
Then, we add up all the x-components and all the y-components
                              x-component (sum) = 0.83
                              y-component (sum) = 2.83 

Then, we determine the magnitude of the block by the equation below.
                             m = √(x² + y²)  = √(0.83² + 2.83²) ≈ 2.95 blocks

Thus, the displacement from the original position is approximately equal to 2.95 blocks. 

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