# Arvin the ant is on a picnic table. He travels 30cm eastward, then 25cm northward, and finally 15cm westward.what is the magnitude of arvin's net displacement?

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Let the starting point be O and E , N an W be the three postion the ant moves, Then OE=30cm east , EN = 25 cm and NW = 30.

Therefore, if we take the OX towards east as X coordinate, and OY towards north as Y cordinates, then the coordinates of the postion of O is (0,0), E is ( 30, 0), N is (30 , 25) and W = ( 30-15 , 25) = (15, 25)

Therefore, the final postion of W =(15,25)

The distance,d between two points (x1,y1) and (x2,y2) is given by d= sqrt[(x2-x1)^2+(y2-y1)^2]

Therefore, OW = sqrt[(15-0)^2+(25-0^2)]= sqrt(225+625)=sqrt(850)= 5sqrt34

With the first movement of 30 cm east the net displacement is 30 cm to east.

With the second movement of 25 cm north the net displacement is 30 cm to east and 25 cm to north.

With final movement of 15 cm to the west, there is no change in the northward displacement, which remains 25 cm. The eastward displacement gets reduced by 15 cm. So east ward displacement is 30 - 15 = 15 cm.

So the net displacement is 25 cm to north and 15 cm to east.

The magnitude of this displacement is

= [(Displacement to north)^2 + (Displacement to east)^2]^(1/2)

= [25^2 + 15^2]^(1/2) = (625 + 225)^(1/2) = 850^(1/2) = 29.155 cm (approximately)

Answer: Magnitude of Arvin's net displacement is 29.155 cm.