AP Calculus AB · Topic 8.2

Position, Velocity, and Acceleration using Integrals Practice

Part of Applications of Integration.(CHA-4.B)

Practice questions

13

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Sample questions

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  1. Sample 1difficulty 2/5

    Average velocity of v(t)=t22tv(t) = t^2 - 2t on [0,3][0, 3] is

    • A

      32\tfrac{3}{2}

    • B

      00

      check_circle
    • C

      11

    • D

      33

    Why

    1303(t22t)dt=(1/3)(99)=0\dfrac{1}{3}\int_0^3 (t^2 - 2t)\,dt = (1/3)(9 - 9) = 0.

  2. Sample 2difficulty 2/5

    A particle moves with v(t)=costv(t) = \cos t on [0,2π][0, 2\pi]. Total displacement:

    • A

      44

    • B

      2π2\pi

    • C

      22

    • D

      00

      check_circle

    Why

    02πcostdt=0\int_0^{2\pi}\cos t\,dt = 0.

  3. Sample 3difficulty 2/5

    For a particle with v(t)=2tv(t) = 2t, the displacement over [0,3][0, 3] is

    • A

      1818

    • B

      99

      check_circle
    • C

      66

    • D

      33

    Why

    032tdt=[t2]03=9\int_0^3 2t\,dt = [t^2]_0^3 = 9.

  4. Sample 4difficulty 2/5

    With v(t)=2t+1v(t) = 2t + 1 and s(0)=4s(0) = 4, find s(3)s(3).

    • A

      1313

    • B

      1616

      check_circle
    • C

      1010

    • D

      2222

    Why

    s(t)=t2+t+Cs(t) = t^2 + t + C. s(0)=4=Cs(0) = 4 = C. s(3)=9+3+4=16s(3) = 9 + 3 + 4 = 16.

  5. Sample 5difficulty 2/5

    A particle starts at s(0)=5s(0) = 5 with velocity v(t)=3tv(t) = 3t. Position at t=4t = 4:

    • A

      2424

    • B

      2929

      check_circle
    • C

      4848

    • D

      1212

    Why

    s(t)=5+0t3udu=5+3t2/2s(t) = 5 + \int_0^t 3u\,du = 5 + 3t^2/2. s(4)=5+24=29s(4) = 5 + 24 = 29.