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6 March, 12:07

The x-coordinates of two objects moving along the x-axis are given as a function of time t. x1 = (4 m/s) t and x2 = - (159 m) + (24 m/s) t - (1 m/s2) t2. Calculate the magnitude of the distance of closest approach of the two objects. x1 and x2 never have the same value.

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  1. 6 March, 12:18
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    The x-coordinate of the first object is

    x₁ (t) = (4 m/s) t

    The x-coordinate of the second object is

    x₂ (t) = - (159 m) + (24 m/s) t - (1 m/s²) t²

    The distance between the two objects is

    x (t) = x₂ - x₁

    = - 159 + 24t - t² - 4t

    = - t² + 20t - 159

    Write this equation in the standard form for a parabola.

    x = - [t² - 20t] - 159

    = - [ (t - 10) ² - 100] - 159

    = - (t-10) ² - 59

    This parabola has a vertex at (-10, - 59), and it is downward.

    Because the maximum value of x is negative, the two objects never touch

    The closest distance between the objects is 59 m.

    The two graphs confirm that the analysis is correct.

    Answer: The closest approach is 59 m.
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