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

Question

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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Thông Đạt 1 year 2021-09-01T21:28:12+00:00 2 Answers 14 views 0

Answers ( )

    0
    2021-09-01T21:29:14+00:00

    Answer:

    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

    0
    2021-09-01T21:30:01+00:00

    Answer:

    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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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )