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If a particle is released from rest at 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, and What is the meaning of the negative root of this quadratic equation?

If a particle is released from rest  at 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, and  What is the meaning of the negative root of this quadratic equation?

Grade:11

2 Answers

Jitender Pal
askIITians Faculty 365 Points
9 years ago
From equation 2-28, we have
x = x_{0} + v_{0x}t + \frac{1}{2}a_{x}t^{2} …… (1)

If the particle is at rest (v0x = 0 ) at x = 0, at the time t = 0 , the equation (1) can be written as:
x = \frac{1}{2} a_{x}t^{2}
t = \pm \sqrt{\frac{2x}{a_{x}}}
It is clear from the equation above that for fixed value of x and ax, 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.
Prachi Priya
13 Points
5 years ago
It means that if another particle matching the size, mass, etc and other properties with that particle then if that another particle would have been thrown upwards at the same time (when the particle was released from rest) it would have reached that point x by the time of negative root of the equation.

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