Going by 12 meters in 0.62 seconds..
Average speed = 12 / 0.62 = 19.35 meters per second.
Assuming you start from 0 meters per second, and assuming linear acceleration (unrealistic, will come it in a bit), then you must accelerate up to 19.35 x 2 = 38.7 meters per second.
Accelerating up to 38.7 meters per second in 0.62 seconds, 38.7 / 0.62, gives you an acceleration of 62.41 meters per second per second.
However, it’s much more likely the acceleration will be higher just after the start, and will drop off towards the end, so this is, if you like, somewhat of a lower bound.
62.41 / 9.81 = 6.36 G during acceleration.
Deceleration from 38.7 meters per second over 0.3 meters, again, we’re decelerating to 0 meters per second, so assuming linear deceleration we will have an average speed of 19.35 meters per second again.
Covering 0.3 meters at an average of 19.35 meters per second will take 0.3 / 19.35 = 0.016 seconds.
Decelerating from 38.7 meters per second in 0.016 seconds will be a deceleration of 38.7 / 0.016 = 2418.75 meters per second per second.
2418.75 / 9.81 = 246.6 G during deceleration.
I think that’s all roughly right, it’s been a long time since I’ve done this sort of thing. Assuming linear acceleration does generally give you a lower bound on these things though, so the actual G force could be much higher.
For comparison, say you drive a car into a brick wall (silly, I know) at 40mph, and the front end of the car crumples inwards by 1 meter (hence giving you a stopping distance of 1 meter from 40mph).
That should give you about 16.5 G. Which would hurt.
