Sagot :
DISTANCE AND DISPLACEMENT
==============================
» 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]
==============================
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]
==============================
#CarryOnLearning
(ノ^_^)ノ
