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Displacement, Velocity, and Acceleration

AP Physics 1· difficulty 3/5

A ball dropped from rest in vacuum falls. Compared to the distance fallen during the <strong>first</strong> second, the distance during the <strong>second</strong> second is

  • A

    3×3 \times

    check_circle
  • B

    4×4 \times

  • C

    2×2 \times

  • D

    Same

Explanation

First second: 12g(1)2=5 m\tfrac{1}{2}g(1)^2 = 5~\text{m}. Through 2 s2~\text{s}: 12g(2)2=20 m\tfrac{1}{2}g(2)^2 = 20~\text{m}. Second-second distance: 20−5=15 m20 - 5 = 15~\text{m}, which is 3×3\times the first.

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