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13 September, 12:01

The front 1.20 m of a 1,550-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a car traveling 24.0 m/s stops uniformly in 1.20 m, how long does the collision last

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Answers (2)
  1. 13 September, 12:19
    0
    t = 0.1 s

    Explanation:

    Given:

    - The distance crushed (Stopping-Distance) s = 1.20 m

    - The mass of the car m = 1,550 kg

    - The initial velocity vi = 24.0 m/s

    Find:

    - How long does the collision last t?

    Solution:

    - The stopping distance s is the average velocity v times time t as follows:

    s = t * (vf + vi) / 2

    Where,

    vf = 0 m/s (Stopped)

    s = t * (vi) / 2

    t = 2*s / vi

    t = 2*1.20 / 24

    t = 0.1 s
  2. 13 September, 12:29
    0
    t=0.1seconds.

    Explanation:

    The mass of the car m = 1,550 kg

    We know the stopping distance, 1.20m,

    we know the final velocity, 0m/s (its stopped),

    the starting velocity, 24m/s.

    d=t (v2+v1) / 2

    Rearange to solve for t, and remove the v2^2 as its zero

    t=2d/v1

    t=2 (1.20m) / (24m/s)

    t=0.1seconds.
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