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Subtracting Vectors Graphically: Example Problem Tallula cycles with a velocity of 5.0 [N]. She then turns a corner and cycl
Tallula walks 5.00 m 38° N of E. She then walks 8.00 m 25º E of S. What is her total displacement?
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Answer #1

1)
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Consider \Delta V as the change in velocity.
\Delta V = SQRT[v2 + v2]
= SQRT[52 + 52]
= SQRT[50]
= 7.1 m/s
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2)
Consider that north is along y-axis and east is along
Initial displacement, d1 = 5 * cos(38) \hat{x} + 5 * sin(38) \hat{y}
Final displacement, d2 = 8 * sin(25) \hat{x} - 8 * cos(25) \hat{y}

Total displacement, d = d1 + d2
= [5 * cos(38) + 8 * sin(25)] \hat{x} + [5 * sin(38) - 8 * cos(25)] \hat{y}
= 7.32 \hat{x} - 4.17 \hat{y}
|d| = SQRT(7.32)2 + (-4.17)2]
= SQRT[71]
= 8.43 m

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