Electronic – DC Motor very high acceleration

When the motor is spinning it generates a voltage proportional to its rotational speed (almost equal to the supply voltage when running free). Then when you apply a short the current is determined by that voltage and the motor's internal resistance. Torque is proportional to current, so initially you get a braking force equal to the stall torque. However as the motor slows down it produces less voltage, so the current and torque reduces (down to zero when it stops). If the 'short' is not a low resistance compared to the motor's internal resistance then it will have even less braking force.

Any inertia in the drive chain will make it run on past the point where you try to stop. A motor in a gearbox has a lot of inertia due to the high rotational speed of the armature and first gear stages. It will never stop instantly, just like it won't go from stationary to full speed instantly when powered up.

Motors such as the RF300 and RF370 typically have high internal resistance and low torque, relying on the gearbox to provide sufficient output torque. They also have heavy iron-cored armatures which increase inertia. Swapping out the motor for a more powerful coreless type with low internal resistance would improve the braking speed. Unfortunately good coreless motors tend to be expensive, and often require matching (expensive!) gearboxes.

You can stop faster by applying reverse voltage, but be careful because that can cause the transistor bridge to 'shoot-through' if you don't stop and wait for the inductive back-emf to die down before reversing. Also the peak current will be twice as high as normal.

Even with reverse voltage applied it will take some time to stop. To compensate for this you must start braking the motor before the robot gets to the position you want. How much before depends on the amount of inertia in the system, which may vary depending on what the robot is doing.

Looking at the datasheet, while it doesn't lie, it is definitely on the edge of misleading.

Note that the "max power" stated occurs at almost exactly half the unloaded RPM and half the stall current. This is indeed the "max power" point for such a motor, but the datasheet fails to mention that it is also the nominal 50% efficiency point, thus dissipating 12V*68A-337W = 479W in that tiny motor - destroying it, probably in minutes.

(Ideally, exactly half the power would be delivered, about 400W shaft and 400W heat, but the motor isn't ideal).

The motor is probably suitable for 100-150W continuous output and 200-250W short term.

So practically you must operate the motor at the upper end of the speed range, and if the speed falls below (say) 70% of the unloaded speed (or the current rises to 30% of the stall current) then - unless this is strictly temporary, like starting a heavy load or hitting a chilled spot while machining a cast iron surface, you need to cut the current and protect the motor.

Then the question of which side of the torque speed curve doesn't apply - unless the protection has tripped, you should be on the high speed side.

You can get circuit breakers that will allow short-term overcurrent. These are "motor rated" or Class C breakers for the AC motors used in most machine tools. I don't know of anything suitable for 12V DC though. I'd be looking for a 12V DC supply that can be set to trip if its output exceeds 40A for more than a couple of seconds. And as Olin says, if you want to monitor it yourself, measuring the current is definitely the way to go.