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?

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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