Meaning of negative time Whenever we graph time along a $x$-axis, we graph it along negative and positive $x$-axis both. So what does a negative time mean in the negative case axis? What does the negative sign indicate?
 A: In the simplest possible terms, negative time means last year, last month, last week, yesterday, one hour ago, one minute ago, one second ago. It refers to the past.
A: Suppose you work for NASA and are the person who is given the task of announcing the time to the launch of a particular rocket. You will call this time $t=0$, and so five seconds before the launch you will announce "t minus 5..4..3..2..1..ignition". Now this time $t=0$ could have been for example 28th February 12 noon, and everything before that particular event exists on the "negative time" axis, or all points in time before this event.
We do the same thing always in classical mechanics. We can define any point to be $t=0$ and it simply defines what will happen/happened at this point, and everything prior to this point is the past and everything after it is the future.
Consider also a case where we have a position versus time graph, which shows an object with displacement according to the equation $$x(t)=(5t^2+1)\,\mathrm{m},$$ where $\mathrm{m}$ represents metres and $t$ is measured in seconds. You are told its displacement at say $t=1\,\mathrm{s}$  is $6\,\mathrm{m}$. You can confirm this by substituting $t=1\,\mathrm{s}$ into this equation. But now you want to know what was its displacement at $t=-2\,\mathrm{s}$. You can once again substitute this value for $t$ into the same equation and you will know that the object had a displacement of $21\,\mathrm{m}$, two seconds before it reached the axis defined by $t=0\,\mathrm{s}$.
"Negative time" simply means all points in time before a specific event, that we say will happen at $t=0\,\mathrm{s}$.
We do a very similar thing in relativity with spacetime diagrams. In such cases, the spacetime origin is represented by $x = 0$ and $t = 0$ and represents the present time and location of the observer (in that reference frame). Events with $t > 0$ are in the future, and events with $t< 0$ are in the past of this observer.
We can choose the location of the origin to make the solution of a problem as
simple and convenient as possible when dealing with regular dynamics.
A: The convention is that positive time is the future and negative time is the past. If you assume this convention then you can explicitly show that Newton's laws are time reversible or time symmetric.
A ball is thrown.
$y=v_{oy}t  - \frac{g}{2} t^2$
$x=v_{ox}t$
A film of the thrown ball is played in reverse. $t \to -t$
$y=-v_{oy}t  - \frac{g}{2} t^2$
$x=-v_{ox}t$
In both cases, $m\frac{dy^2}{dt^2}=-mg$.
If you watch a film of a ball moving along such a path then you can't tell whether the film is going forward in time or backward. Both motions are consistent with Newton's Laws and nothing looks unusual if time is reversed.
So, yes, it is ultimately a convention, but it is a convention that works. Obviously, you could completely reverse the convention and that would be consistent with Newton's Laws.
