Total stopping distance = Reaction distance + Braking distance

During reaction time ($t_r$): driver doesn't brake, car moves at speed v.

• Reaction distance: $d_r$ = v * $t_r$

During braking (deceleration a):

• $v^2$ = 2a$d_b$ => $d_b$ = v^$\frac{2}{2a}$

Total: d = v*$t_r$ + v^$\frac{2}{2a}$

Key insight: Braking distance is proportional to $v^2$. Doubling speed quadruples braking distance.

JEE application: If a car at 20 m/s stops in 50 m, find the deceleration. 0 = (20)^2 + 2a50 => a = -400/100 = -4 m/$s^2$