If a particle is released from restat x0 = 0 at the time t = 0, Eq. 2-28 for constant acceleration says that it is at position x at two different times-namely, andWhat is the meaning of the negative root of this quadratic equation?

From equation 2-28, we have
…… (1)

If the particle is at rest (v_0x = 0 ) at x = 0, at the time t = 0 , the equation (1) can be written as:


It is clear from the equation above that for fixed value of x and a_x, we get two solutions for time. It is meaningless to define the state of the object before the beginning of observation. Henceforth the solution corresponding to the negative time in the quadratic equation above is discarded.