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Displacement, Velocity, and Acceleration

AP Physics 1· difficulty 2/5

A car decelerating uniformly takes distance dd to stop from speed vv. From speed 2v2v on the same surface, the stopping distance is

  • A

    3d3d

  • B

    dd

  • C

    4d4d

    check_circle
  • D

    2d2d

Explanation

With v2=v02+2adv^2 = v_0^2 + 2 a d and v=0v = 0, d∝v02d \propto v_0^2. Doubling speed quadruples stopping distance.

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AP Physics 1 · Displacement, Velocity, and Acceleration Practice Question | Acemy