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Representing Motion

AP Physics 1· difficulty 2/5

0 3 9 +4 0 −2 t (s) a (m/s²)

An object starts from rest at t=0t = 0. Its acceleration is +4 m/s2+4~\text{m/s}^2 for the first 3 s3~\text{s}, then −2 m/s2-2~\text{m/s}^2 from t=3 st = 3~\text{s} to t=9 st = 9~\text{s}. What is its velocity at t=9 st = 9~\text{s}?

  • A

    −24 m/s-24~\text{m/s}

  • B

    +6 m/s+6~\text{m/s}

  • C

    +12 m/s+12~\text{m/s}

  • D

    00

    check_circle

Explanation

Δv=∫a dt=(+4)(3)+(−2)(6)=12−12=0\Delta v = \int a\,dt = (+4)(3) + (-2)(6) = 12 - 12 = 0. Starting from rest, final velocity is 00.

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