The human body can survive an acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than [closed]

The human body can survive an acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than 250 m/s². If you are in an auto- mobile accident with an initial speed of 105 km/h and you are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?

So, I modeled the problem with this diagram: It says that the initial position is $$0m$$ and initial velocity is $$29.2\frac{m}{s}$$. After $$t'$$ time has passed, the car crashes; it has traveled a distance of $$X$$, and since it crashed, velocity is 0.

Acceleration is fixed at its maximum value, which is $$250 \frac{m}{s²}$$

Now, I use this formula to obtain $$X$$: $$v_x^2=v_{0x}^2+2a_x\left(x-x_0\right)$$

replacing, I get

$$0,00\,\frac{m}{s} = 852.6\,\frac{m²}{s²}+500\,\frac{m}{s²}X$$ so I solve for X:

$$X = \frac{-852.6\,\frac{m²}{s²}}{500\,\frac{m}{s²}} = -1.70\,m$$

But it's kind of weird that I get a negative distance, so I check the solutions manual; what I get is that $$1.70$$ is correct, but direction is wrong (the negative sign): The reason is that they choose a negative acceleration, but I don't get why. Could you explain to me why it has to be chosen negative?

closed as off-topic by ZeroTheHero, Ben Crowell, Aaron Stevens, John Rennie, Jon CusterJan 18 at 14:21

This question appears to be off-topic. The users who voted to close gave this specific reason:

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• Don't get all "hung up" on the negative sign. This is physics, not math. – David White Jan 18 at 1:31
• I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. – Ben Crowell Jan 18 at 4:12