>same intensity and deceleration.

It cannot be both. It mathematically cannot be both. They can brake at the same rate (acceleration) or intensity (conversion of kinetic energy into heat) but because they are traveling different speeds those two values cannot be the same for both cars.

The math you did was for intensity, not force/acceleration, which because of the ^2 in the KE equation exaggerates the difference. Whereas if you did the math based on force you'd get a mild, linear, difference.

> and braked at the same rate,

You're being a bit sly with word choice here. You're doing the math for conversion of KE into heat whereas in common parlance "rate" means force/acceleration.

Braking "at the same rate" [of energy conversion] is way less actual braking force for the faster car.

This is basically the same kinetic energy into heat math wherein you can descend a grade at a low speed, apply a force and be fine and descend the same grade at a higher speed and apply the same force and cook the brakes. Or you can apply less force, and get the same amount of energy conversion into heat (i.e. your wording trick in the proposed scenario)

You've taken what's basically the math behind trucks descending a grade (rate of energy conversion is actually limited by ability of brakes to shed heat, not friction) and re-framed it as cars stopping to create a trick question.

OP wasn't explicit about taking the work = force * distance approach to dissipating energy. Two cars with the same mass and braking force (and thus deceleration) will put the same amount of work into the vehicle per unit distance, so will dissipate the same amount of energy in the braking maneuver.

You are right that the faster car is converting kinetic energy into heat faster per unit time. It also has less time to do so. The work formulation of the problem makes it obvious that these have to cancel out exactly.