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You first went to the bench, and then walked 2 meters North to reach the seesaw. After that, you again
walked 5 meters to the East to play in the swing. Then you saw a beautiful flowering plant, so you
walked 2 meters South to the flowering plant. You got tired and walked 5 meters to the West so that
you can sit in the bench.
Q1. What is the distance (how far was the path traveled) covered during your entire walk?
Q2. What is the displacement (change in position) during your entire walk?​


Sagot :

DISTANCE AND DISPLACEMENT

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» Q1. What is the distance (how far was the path traveled) covered during your entire walk?

» To find the distance, just add all the distances where he travelled.

[tex] \implies \sf \large 2m + 5m + 2m + 5m = 14m[/tex]

Final Answer:

[tex] \tt \huge » \: \purple{14 \: meters}[/tex]

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Q2. What is the displacement (change in position) during your entire walk?

» When finding the displacement, we were using coordinates and the Pythagorean Theorem.

» In coordinates, (x, y) is common, by using West and East, we use (x) coordinates. When going East (add), when going to West (subtract).

» For North and South, we use (y) coordinates. Going North (add), South (subtract).

» The initial position will be (0, 0)

  • › Going North 2 meters (0, 0+2) = (0, 2)
  • › Going East 5 meters (0+5, 2) = (5, 2)
  • › Going South 2 meters (5, 2-2) = (5, 0)
  • › Going West 5 meters (5-5, 0) = (0, 0)

» Now the Pythagorean Theorem comes in, the (x) and (y) indicates as the leg of a right triangle while the hypotenuse is the displacement. Using the formula we can identify the displacement.

[tex] \: : \implies \sf \large {x}^{2} + {y}^{2} = {d}^{2} [/tex]

[tex] \implies \sf \large {0}^{2} + {0}^{2} = {d}^{2} [/tex]

[tex] \implies \sf \large 0 = {d}^{2} [/tex]

[tex] \implies \sf \large \sqrt{0} = \sqrt{ {d}^{2} } [/tex]

[tex]\implies \sf \large d = 0[/tex]

Final Answer:

[tex]\tt \huge » \: \purple{0 \: meter}[/tex]

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