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R2=5i,10j, 20k
ditanya delta R
jawab :
delta r= r2-r1
delta r=(5-10)i+(10-5)j+(20-15)k
delta r = -5j+5i+5k
Besarnya = √[(-5)² + 5² + 5²]
= satuan
2. Posisi awal: r(0) = 0 i + 0 j
Posisi akhir: r(t) = 100 i - 50 j (di sini dianggap timur adalah x+ dan selatan y-)
Kecepatan rata-rata = Perubahan posisi / selang waktu = 100 i - 50 j km/jam
Semoga Bermanfaat ^_^