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Position, Velocity, and Acceleration using Integrals

AP Calculus AB· difficulty 4/5

With v(t)=3t2−12t+9v(t) = 3t^2 - 12t + 9 on [0,4][0, 4], the total distance traveled by the particle is

  • A

    44 (this is the net displacement, not the total distance)

  • B

    00

  • C

    1212

    check_circle
  • D

    88

Explanation

v=0v=0 at t=1,3t=1,3. Position s(t)=t3−6t2+9ts(t)=t^3-6t^2+9t gives s(0)=0, s(1)=4, s(3)=0, s(4)=4s(0)=0,\,s(1)=4,\,s(3)=0,\,s(4)=4. Total distance = ∣4−0∣+∣0−4∣+∣4−0∣=12|4-0|+|0-4|+|4-0| = 12. (Net displacement is only s(4)−s(0)=4s(4)-s(0)=4.)

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