Position, Velocity, and Acceleration using Integrals

AP Calculus AB· difficulty 3/5

For v(t)=2t6v(t) = 2t - 6 on [0,4][0, 4], the <strong>total distance</strong> traveled is

  • A

    1010

    check_circle
  • B

    00

  • C

    1212

  • D

    2020

Explanation

v|v|: v<0v < 0 on [0,3][0, 3], v>0v > 0 on [3,4][3, 4]. Distance = 03(2t6)dt+34(2t6)dt=9+1=10\int_0^3 -(2t-6)\,dt + \int_3^4 (2t-6)\,dt = 9 + 1 = 10.

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