Child's Displacement on Cubical Jungle Gym: Solving the Problem

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SUMMARY

The problem involves calculating the displacement of a child climbing from one corner of a cubical jungle gym to the diagonally opposite corner. The jungle gym has a side length of 2 meters. To solve this, the displacement can be broken down into two segments: first, the horizontal distance across the base (2m on both x and y axes), and then the vertical distance (2m on the z-axis). Using the Pythagorean theorem, the total displacement can be calculated as √((2² + 2²) + 2²) = √(12) = 2√3 meters.

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I am having problems figuring this out. How should I go about this?


A child starts at one corner of a cubical jungle gym in a playground and climbs up to the diagonally opposite corner. The original corner is the coordinate origin, and the x-, y- and z-axes are oriented along the jungle gym edges. The length of each side is 2m. The child's displacement is?
 
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Break it down into two smaller and easier problems. And draw yourself a diagram! Solve using Pythagoras a² + b² = c²
 

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