Selasa, 13 Oktober 2009

JAWABAN UJI KOMPETENSI KINEMATIKA GERAK LURUS

KELAS : XI IPA 1

KODE : A

  1. dan (5)

    r = (6t2 – 2t – 5)i + (3t2 – 5t + 1)j

    1. v = dr/dt = (12t – 2)i + (6t – 5)j

      t = 1 ® v = 10i + j

      v = Ö101 m/s

    2. a = dv/dt = 12i + 6j

      a = Ö180

  2. dan (6)

    a = 4i + 6j

    v0 = i +3j

    v = v0 + ò a .dt

    = (i + 3j) + (4t i + 6t j)

    = (4t +1) i + (6t+3)j

    = 13i + 21j

    v = Ö610 m/s

  3. dan (7)

    r = (4t – 2) i + (2 – 3t) j

    1. t1 = 2 Þ r1 = 6 i – 4 j m

      t2 = 4 Þ r2 = 14 i – 10 j m

      vr = {(14 i – 10 j) – (6 i – 4 j)}/2

      = 4 i – 3 j m/s

    2. v = dr/dt = 4 i – 3 j m/s
  4. dan (8)

    r0 = 3 i + 4 j

    v = (4t + 3) i + (8t – 6) j

    1. r = r0 + ò v dt

      = (3 i + 4 j) + ò(4t + 3) i + (8t – 6) j dt

      = (3 i + 4 j) + (2t2 + 3t) i + (4t2 – 6t) j

      = (2t2 + 3t + 3) i + (4t2 – 6t + 4) j

      = (2.22 + 3.2 + 3) i + (4.22 – 6.2 + 4) j

      = 17 i + 8 j meter

    2. a = dv/dt = 4 i + 8 j m/s2


 


 


 


 


 


 

KODE B :

  1. dan (5)

    r0 = i + 5 j

    v = (6t -2) i + (8 – 2t) j

    1. a = dv/dt = 6 i – 2j
    2. r = r0 + ò v dt

      = (i + 5j) + ò(6t – 2) i + (8 – 2t) j dt

      = (i + 5j) + (3t2 – 2t) i + (8t – t2) j

      = (3t2 – 2t + 1) i + (8t – t2 + 5) j

      = 2 i + 12 j

  2. dan (6)

    r = (4 – 6t) i + (4 – 2t) j

    1. t1 = 2 Þ r1 = (4 – 6.2) i + (4 – 2.2) j = - 8 i

      t2 = 4 Þ r2 = (4 – 6.4) i + (4 – 2.4) j = - 20 i – 4 j

      vr = {(- 20 i – 4 j) – (-8 i)}/2 = - 6 i – 2 j

    2. v = dr/dt = -6 i – 2 j
  3. dan (7)

    r = (5t2 – 7t + 4) i + (2t2 – 5t – 2) j

    1. v = dr/dt = 23 i + 7 j

      = Ö578 m/s

    2. a = dv/dt = 10 i + 4 j

      a = Ö116 m/s2

  4. dan (8)

    v0 = 5 i + j

    a = 7 i + 5 j

    v = v0 + ò a . dt

    = (5 + 7t) i + (1 + 5t) j

    = (5 + 7.2) i + (1 + 5.2) j

    = 19 i + 11 j

    v = Ö482 m/s

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