kwikee
Likes Dirt
Maybe.To brake, a heavier rider has to use more force at the levers, but his grip ( ie distance to stop) is the same because grip is a function of weight times co-efficient of friction ; the light rider carries less energy and therefore uses less energy to stop, but they both stop in the same distance.
Make the assumption that the 2 theoretical riders are using well bedded in Saints, and could lock the front wheel with maximum brake lever application, so they will modulate their braking to the maximum available traction to achieve the shortest braking distance.
The formula for kinetic energy (energy of motion) is E(kinetic)=1\2 mass x Velocity squared.
So also assume velocity is the same at the start of braking, then the mass is the only other factor. If mass of rider 1 is greater than the mass of 2, then the kinetic energy carried by 1 will be greater, by a factor of half their difference in mass.
The only other difference will apply where rider 1 is sending more Newtons per square mm of force through the contact patch of their tyres, than rider 2, and therefore can brake harder without locking up, providing greater resistance, and shedding slightly more kinetic energy per second.
Whether this equals the same distance would need to be calculated but it stands to reason.