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physics.fisikastudycenter.com,- Kinematics and kinematics with vectors analysis, enrichment problems for high school students grade 11, problem 1. Id: Kinematika dan Kinematika dengan analisis vektor Kelas 11 SMA, contoh soal dan pembahasan jawaban.

Problem 1
A particle with electric charge in it travels in a straight motion with an initial velocity of 100 m s⁻¹. The particle is accelerated by an electrical force with an acceleration of a = (2 – 10t) m s⁻² where t is the time when the force is working. Find the the velocity when t = 4 sec!
→ Sebuah partikel bermuatan listrik mula-mula bergerak lurus dengan kecepatan 100 m s⁻¹. Karena pengaruh ga ya listrik, partikel mengalami percepatan yang dinyatakan dengan persamaan a = (2 – 10t) m s⁻² (t adalah waktu lamanya gaya listrik bekerja). Kecepatan partikel setelah gaya bekerja selama 4 sekon adalah …
A. 24 m s⁻¹
B. 28 m s⁻¹
C. 32 m s⁻¹
D. 36 m s⁻¹
E. 40 m s⁻¹
Question source: EBTANAS Fisika 1997

Solution

ν = ν_o + ₀∫⁴ a dt
ν = 100 + ₀∫⁴ (2−10t) dt
ν = 100 + [2t−5t²]₀⁴
ν = 100 + [2(4)−5(4)²] = 100 + (8-80) = 100−72 = 28 28 m s⁻¹
Answer : B

Prepared by fisikastudycenter.com
Sources/Literatures
-Soal Ebtanas 2001

Problem 2
A particle when t₁ = 0 is at (2, 4) and t₂ = 2 sec at (8, 6) then its average velocity vector is....
→ Sebuah partikel pada t₁ = 0 berada pada koordinat (2, 4) dan pada t₂ = 2 detik berada pada koordinat (8, 6) maka vektor kecepatan rata-ratanya adalah ....
A. 3i + 2j
B. 4i + 3j
C. 3i + j
D. 2i + 4j
E. 4i + 3j
Question source: EBTANAS Fisika


Solution

Particle position when t₁ = 0 in i, j
r₁ = 2i + 4j

Particle position when t₂ = 2 in i, j
r₂ = 8i + 6j

Using definiton of average velocity then we get



Answer : C

Prepared by fisikastudycenter.com
Sources/Literatures
-Soal-Soal Ebtanas Fisika

Problem 3
The equation of a particle position : r = {(15t√3) i + (15t−5t²) j } m. When t = 1.5 sec the particle speed is....
→ Posisi sebuah benda dinyatakan dengan persamaan r = {(15t√3) i + (15t−5t²) j } m. Setelah benda bergerak selama 1,5 sekon kelajuannya menjadi....
A. 0
B. 15 m s⁻¹
C. 11,5√3 m s⁻¹
D. 22,5 m s⁻¹
E. 15√3 m s⁻¹
Question source: EBTANAS Fisika 2002


Solution

From the equation of position we could get speed equation, still in i, j



Finally the magnitude of speed without i and j,



Prepared by fisikastudycenter.com
Sources/Literatures
-Soal-Soal Ebtanas Fisika

Problem 4
A particle travels with an equation of y = 30 t – t² m. If y is the vertical direction, find the maximum height reached by this particle!
→ Sebuah benda bergerak dengan persamaan y = 30 t – t² meter. Jika y adalah arah vertikal. Maka ketinggian maksimum benda tersebut adalah...
A. 55 m
B. 75 m
C. 125 m
D. 175
E. 225 m
Question source: Modification of Soal UAN Fisika SMA 2004 Fisika Study Center


Solution

When the particle reaches the highest point, its velocity is 0 m/sec. We could find the velocity from the position equation, get the time when v = 0 sec, and insert back to y equation as below steps



Prepared by fisikastudycenter.com
Sources/Literatures
-Soal-Soal Ebtanas Fisika

Problem 5
A block of 2 kg of mass is moved horizontally at a frictionless table from a rest by a force F where F = 60 + 3t with t is in sec unit and F is in N unit. Find:
a) the acceleration of the block for t = 2 sec
b) the speed of the block for t = 2 sec
→ Sebuah benda bermassa 2 kg digerakkan mendatar di meja licin dari keadaaan diam oleh sebuah gaya mendatar F yang berubah terhadap waktu menurut F = 60 + 3t dengan t dalam s dan F dalam N. Tentukan :
a) percepatan benda untuk nilai t = 2 s
b) kecepatan benda untuk nilai t = 2 s
Question source: Modification of Soal UMPTN

Solution

F = 60 + 3t
a) a = F/m = (60+3t)/2 = 30 t + (3/2)t
t = 2 s ⇒ a = 30(2) + (3/2)(2) = 30 + 3 = 33 m/s²
b) ν = ν_o + ∫adt
ν = [30t + (3/4)t²]
t = 2 s ⇒ ν = 30(2) + (3/4) (2)² = 63 m/s

Prepared by fisikastudycenter.com
Sources/Literatures
-Soal-Soal UMPTN/SPMB