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# Answer to Question #105768 in Mechanics | Relativity for Sommy

Question #105768
A man walks 1km due east and then 1km due North his displacement
1
2020-03-19T11:22:09-0400

The movement of a person to the East and then to the North corresponds to his movement along the sides of a right triangle. In this case, the path that it has passed is equal to the sum of the cathetus, and its offset is the length of the hypotenuse.

Let's the initial point is A=(0,0) then the final point will be B=(1,1). Thus the displacement is "\\vec {AB}=B-A=1\\hat i+1\\hat j- (0\\hat i+0\\hat j)=(1-0)\\hat i+(1-0)\\hat j=(1,1)"

Using the Pythagorean law for right triangles we get "AB=\\sqrt{1km^2+1km^2}=\\sqrt{2km^2}=1.41 km"

Answer: The displacement of a man from the initial point walk is (1,1) with the length

1.41 km

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