Travelled displacement - how do we take care of the possibility that it might be negative?

so this is a very basic question but I can't really find a full answer.

So let's assume we have a point of mass $m$ travelling along the x-axis with a acceleration a. The travelled displacement is:

$$\Delta x = x(t_E) - x(t_B) = \frac{1}{2} a (t_E^2 - t_B^2)$$

Now, that makes sense, but what if we travel backwards, such that $a<0$? Then we would get $\Delta x < 0$ which isn't possible since the displacement is the difference between two points i.e. a relative quantity.

I'm now wondering how that is solved. Is it just by argument i.e. "we just take the absolute value of $\Delta x$" or is my math wrong? Or do actually just use the following formula?

$$\Delta x = x(t_E) - x(t_B) = \frac{1}{2} |a| (t_E^2 - t_B^2)$$

How do we deal with the possibility that the displacement can be negative?

How do we deal with the possibility that the displacement can be negative?

There's no reason to think that displacement can't be negative.

If your $\mathbf{\hat{x}}$ vector points from left to right, then a positive displacement means the final location is to the right of the initial position. And a negative displacement means the final location is to the left of the initial position.

The formula you have used is for displacement. You can’t use the next formula either.Consider particle moving with some initial velocity along the positive x-axis.The formula you gave or the kinematic equation s=ut+(1/2)at^2 doesn’t work after any time greater than time taken to revert it’s motion. But the formula you gave for distance works only when particle starts from rest and moves with uniform acceleration or deceleration.

• so basically I have to argue by myself depending on the situation? – handy Aug 13 '18 at 15:03
• Yes, why do you always need formulas to solve? It’s good to analyse the situation sometimes and find the solution from the basic formulae. – user167540 Aug 13 '18 at 15:13
• I'm currently "cleaning up my knowledge" - like revieweing it. That's all. :) And I just like to be very pedantic with my mathematics. – handy Aug 13 '18 at 15:23
• Hey, why did u make the distance as displacement – user167540 Aug 13 '18 at 17:07

"Then we would get Δx<0 which isn't possible since the displacement is the difference between two points i.e. a relative quantity."

There's something significant about the displacement that you seem to be missing here. Displacement is a vector. It measures both the magnitude and direction of the change in location.

If you're only considering the one dimensional case of the x-axis, as you are here, then there are really only two important directions. The $+x$ direction; which you likely defined as "to the right" and the $-x$ direction, which you would define as the opposite of $+x$, or the left. To really allow for self-consistency, and to use the displacement as a vector, you actually need to define the "+" and "-" directions when specifying a displacement (though no sign is interpreted as positive as well).