Position, Velocity, and Acceleration using Integrals

AP Calculus AB· difficulty 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

Explanation

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

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