Position, Velocity, and Acceleration using Integrals

AP Calculus AB· difficulty 3/5

t v 0 6 5

The distance traveled from t=0t=0 to t=6t=6 at constant velocity v=5v=5 is:

  • A

    55

  • B

    66

  • C

    1111

  • D

    3030

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Explanation

Distance = area under v-t graph =5×6=30= 5 \times 6 = 30.

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