Tolong donk yg bisa jawab dgn penjelasanx.. cuma nmor 1,2,dan5

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Nomor 1
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Diket : 
Posisi awal, r1 = 10i - 4j
Posisi akhir, r2 = 7i + 3j

Tanya :
Vektor dan besar perpindahan, Δr = __?

Jawab :
Δr = r2 - r1
Δr = [7i + 3j] - [10i - 4j]
Δr = (7 - 10)i + (3 - (- 4))j
Δr = - 3i + 7j

Besar nya
|Δr| = √(- 3)² + 7²
|Δr| = √(9 + 49)
|Δr| = √58 meter


Nomor 2
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Diket : 
Vektor posisi, r = (2t³ - 4t²)i + 3t²j

Tanya : 
Besar dan arah peerpindahan dari t1 = 10 sec sampai t2 = 25 sec
Δr = __? 

Jawab :
Step 1
--------
Cari vektor posisi saat t2 = 25 sec
r(t) = (2t³ - 4t²)i + 3t²j
r(25) = (2.(25)³ - 4.(25)²).i + 3.(25)².j
r(25) = (31250 - 2500).i + (1875).j
r(25) = (28750).i + (1875).j

Step 2
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Cari vektor posisi saat t1 = 10 sec
r(t) = (2t³ - 4t²)i + 3t²j
r(10) = (2.(10)³ - 4.(10)²).i + 3.(10)².j
r(10) = (2000 - 400).i + (300).j
r(10) = (1600).i + (300).j

Step 3
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vektor perpindahan, Δr = __?
Δr = r2 - r1
Δr = r(25) - r(10)
Δr = {(28750).i + (1875).j} - {(1600).i + (300).j}
Δr = (28750 - 1600).i + (1875 - 300).j 
Δr = (27150).i + (1575).j

Besar Perpindahan, |Δr| = __?
|Δr| = √(27150² + 1575²) 
|Δr| = √(737122500 + 2480625) 
|Δr| = √739603125
|Δr| = 27195,65 m

Arah perpindahan nya, θ = __?
tan θ = ry / rx
tan θ = 1575 / 27150
tan θ =  0,0580
θ = arc. tan 0,0580
θ = 3,319° terhadap sumbu x mendatar


Nomor 5
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Diket : 
Jarak, s = 4t³ + 2t² + 8t

Tanya : 
A. Percepatan rata-rata, saat t1 = 0 sampai t2 = 2 sec, a = __?
B. Percepatan sesaat t = 2 sec, a = __?

Jawab :

A] percepatan rata-rata
Step 1
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Cari dulu kecepatan, V = __?
Gunakan rumus diferinsial,
V = ds/dt
V = d[4t³ + 2t² + 8t]/ dt
V = 12.t² + 4.t + 8

Step 2 
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Kecepatan saat t = 2 sec, V2 = V(2) = __?
V(t)  = 12.t² + 4.t + 8
V(2) = 12.(2)² + 4.(2) + 8
V(2) = 48 + 8 + 8
V(2) = 64 m/s

dan 

Kecepatan saat t = 0 sec, V1 = V(0) = __?
V(t)  = 12.t² + 4.t + 8
V(0) = 12.(0)² + 4.(0) + 8
V(0) = 0 + 0 + 8
V(0) = 8 m/s

Step 3
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Percepatan rata-rata, a = __?
a = ΔV/Δt
a = (V2 - V1) / (t2 - t1)
a = (64 - 8) / (2 - 0)
a = (56) / (2)
a = 28 m/s²


B] percepatan sesaat, a = __?
a = dV/dt
a = d[12.t² + 4.t + 8]/dt
a = 24.t + 4

maka besar percepatan sesaat, t = 2 sec adalah 
a = 24.t + 4
a = 24.2 + 4
a = 48 + 4
a = 52 m/s²



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FZA