bantuin jawab no 2 sama 3 kak.. penyelesaiannya gimana?

2. r = 3t i+ 4t j
vektor posisi saat t = 2
r = 3(2) i + 4(2) j
r = 6i + 8j
maka kedudukannya
r = rx i + ry j
rx = 6
ry = 8
(6, 8)

3. r = (2t² + 4t + 2) i + (3t² + 2) j
posisi benda saat t = 0
r = [2(0)² + 4(0) + 2)] i + [3(0)² + 2] j
r0 = 2i + 2j
posisi benda saat t = 1
r = [2(1)² + 4(1) + 2] i + [3(1)² + 2] j
r1 = 8i + 5j
maka perpindahannya
Δr = r1 - r0
Δr = (8i + 5j) - (2i + 2j)
Δr = 6i + 3j