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saat t = 1 ⇒ r1 = (1)² i + 2(1)³ j = i + 2j
saat t = 2 ⇒ r2 = (2)² i + 2(2)³ j = 4i + 16j
∴a). vektor perpindahan Δr = r2 - r1 = (4i + 16j) - (i + 2j) = 3i + 14j (meter)
∴b). besar perpindahan = |Δr| = √[(3)² + (14)²] = √199 meter
c). arah perpindahan, tan θ = [komponen vertikal]/[komponen horisontal]
tan θ = (+14)/(+3) ⇒ KUADRAN 1
∴ arah perpindahan θ = arc tan [14/3] atau tan^-1 [14/3] derajat
ATAU, tan θ = (+14)/(+3) ⇒ tan θ = 14/3 ⇒ θ =77,91° [KALKULATOR]