The SE is about 400 lbs heavier than its ICE counterpart, so it will take longer to stop from the same speed, but nothing alarming.
Not entirely true... The braking force = the tyre coefficient * normal force. So:
Fb = k . Fn
Fn = normal force which is the weight * g, so
Fb = k . mg
The deceleration due to the force is governed by:
F = ma => a = F/m = k.mg/m = kg.
So the deceleration is proportional to the tyre coefficient, and not the weight. *However* in the real world, tyres have a maximum energy they can tolerate, and we have weight transfer. If you put too much energy through the tyre, it will slip as the rubber deforms and/or melts. Weight and/or weight transfer means that the tyre could end up trying to dissipate too much energy and it slips.
So, the ICE car is lighter, which puts less energy through the tyre, however the ICE car has a weight bias to the front (something like 65:35 from memory) whereas the SE is more neutral. This would mean as you brake, you put more stress on the front in the ICE car. So there are 3 likely outcomes:
1) The tyres can handle both the extra weight and extra weight transfer - same stopping distance.
2) The extra weight of the SE means the tyres can't handle as much deceleration - the ICE stops sooner
3) The extra weight over the fronts of the ICE car means the tyres can't handle as much deceleration - the SE stops sooner.
I personally believe 1) to be the correct outcome. With the same tyres and wheels, on the same road, the ICE and SE cars should stop at the same distance.
PS. we ignored ABS, and assume the braking force is enough to lock the tyres if ABS wasn't available. You could also argue that as the brake pads on the SE are always newer, having never been used so they would be more efficient
