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7 August, 18:24

An object moves at 60 m/s in the + x direction. As it passes through the origin it gets a 4.5 m/s^2 acceleration in the - x direction. a) How much time elapses before it returns back to the origin?

b) What is its velocity when it returns to the origin?

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  1. 7 August, 21:25
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    a) After 26.67 seconds it returns back to the origin

    b) Velocity when it returns to the origin = 60 m/s in the - x direction

    Explanation:

    a) Let the starting position be origin and time be t.

    After time t displacement, s = 0 m

    Initial velocity, u = 60 m/s

    Acceleration, a = - 4.5 m/s²

    We have equation of motion s = ut + 0.5 at²

    Substituting

    s = ut + 0.5 at²

    0 = 60 x t + 0.5 x (-4.5) x t²

    2.25t² - 60 t = 0

    t² - 26.67 t = 0

    t (t-26.67) = 0

    t = 0s or t = 26.67 s

    So after 26.67 seconds it returns back to the origin

    b) We have equation of motion v = u + at

    Initial velocity, u = 60 m/s

    Acceleration, a = - 4.5 m/s²

    Time, t = 26.67

    Substituting

    v = 60 - 4.5 x 26.67 = - 60 m/s

    Velocity when it returns to the origin = 60 m/s in the - x direction
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