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v = dr / dt
v = d(2/t i) + 4-1/t^2 / dt ---> turunkan
v = 1i - 1tj ---> masukkan t pada persamaan vektornya
v = 1i - 1(2)j
v = (1xi - 2yj) m/s
besar perpindahan
|r| =√Rx² + Ry²
|r| = √1² + 2²
|r| = √5 m
arah kecepetan partikel
tanФ = y/x = 2/1 = 2
Ф=63 derajat
# untk 1/t = 0,5 t
dan 2/t = 1t