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9 August, 20:36

An egg drops from a second-story window, taking 1.13 s to fall and reaching a speed of 11.1 m/s just before hitting the ground. On contact with the ground, the egg stops completely in 0.140 s. Calculate the average magnitudes of its acceleration while falling and while stopping.

m/s2 (acceleration while falling)

m/s2 (deceleration while stopping)

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  1. 9 August, 21:17
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    While falling, the magnitude of the acceleration of the egg is 9.82 m/s²

    While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²

    Explanation:

    Hi there!

    The equation of velocity of the falling egg is the following:

    v = v0 + a · t

    Where:

    v = velocity at time t.

    v0 = initial velocity.

    a = acceleration.

    t = time

    Let's calculate the acceleration of the egg while falling. Notice that the result should be the acceleration of gravity, ≅ 9.8 m/s².

    v = v0 + a · t

    11.1 m/s = 0 m/s + a · 1.13 s (since the egg is dropped, the initial velocity is zero). Solving for "a":

    11.1 m/s / 1.13 s = a

    a = 9.82 m/s²

    While falling, the magnitude of the acceleration of the egg is 9.82 m/s²

    Now, using the same equation, let's find the acceleration of the egg while stopping. We know that at t = 0.140 s after touching the ground, the velocity of the egg is zero. We also know that the velocity of the egg before hiiting the ground is 11.1 m/s, then, v0 = 11.1 m/s:

    v = v0 + a · t

    0 = 11.1 m/s + a · 0.140 s

    -11.1 m/s / 0.140 s = a

    a = - 79.3 m/s²

    While stopping, the magnitude of the deceleration of the egg is 79.3 m/s²
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