Solving Kinematics w/ Negative Acceleration: Meaning of Negative Root?

[oneplusone]

Suppose a particle is moving along the $x$ axis with velocity $v$. It starts at the point $-5$ and has constant negative acceleration $a$.
You need to find what time the particle reaches the origin.

My teacher solved this question, and used the kinematics equation Δx = v_0t+1/2 at^2

The quadratic had two roots, one of which was negative. What does the negative root mean? Is it extraneous (my teacher just said to ignore it)?

Thanks.

 

[jbriggs444]

The givens of the problem can be seen as specifying the position of the particle not only at all future times, but at all past times as well. The negative solution is the time when the particle was previously at the origin. The positive solution is when the particle will next be at the origin.

 

[oneplusone]

So does this mean that the particle crosses the origin, and later the negative acceleration causes the particle to eventually go backwards, back to the origin?

 

[Nugatory]

So does this mean that the particle crosses the origin, and later the negative acceleration causes the particle to eventually go backwards, back to the origin?

Something is wrong with your signs in the original statement of your problem. If the particle started at -5 (left of the origin) with a positive velocity (right-moving) and a negative acceleration (being pulled to the left) both roots should have been positive, corresponding to exactly what you're describing here.

JBriggs's answer as to the meaning of a negative root in these problems is correct.