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15 May, 19:19

An object starts from rest at time t = 0.00 s and moves in the + x direction with constant acceleration. The object travels 14.0 m from time t = 1.00 s to time t = 2.00 s. What is the acceleration of the object? a) 5.20 m/s^2 b) 10.4 m/s^2 c) 8.67 m/s^2 d) 6.93 m/s^2 e) 12.1 m/s^2

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  1. 15 May, 19:53
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    The acceleration of the object is 9.3 m/s²

    Explanation:

    For a straight movement with constant acceleration, this equation for the position applies:

    x = x0 + v0 t + 1/2 a t²

    where

    x = position at time t

    x0 = initial position

    v0 = initial velocity

    a = acceleration

    t = time

    we have two positions: one at time t = 1 s and one at time t = 2 s. We know that the difference between these positions is 14.0 m. These are the equations we can use to obtain the acceleration:

    x₁ = x0 + v0 t + 1/2 a (1 s) ²

    x₂ = x0 + v0 t + 1/2 a (2 s) ²

    x₂ - x₁ = 14 m

    we know that the object starts from rest, so v0 = 0

    substracting both equations of position we will get:

    x₂ - x₁ = 14

    x0 + v0 t + 1/2 a (2 s) ² - (x0 + v0 t + 1/2 a (1 s) ²) = 14 m

    x0 + v0 t + 2 a s² - x0 - v0 t - 1/2 a s² = 14 m

    2 a s² - 1/2 a s² = 14 m

    3/2 a s² = 14 m

    a = 14 m / (3/2 s²) = 9.3 m/s²
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