CivilForge

Braking Distance

d_{b} = \frac{V ^{2}}{2 g ( f \pm G )}

From v² = 2 a s with a = g(f ± G). V and g must be in consistent units (ft/s and ft/s², or m/s and m/s²). Use +G on upgrades (helps stopping) and −G on downgrades. f is the braking friction coefficient (or a/g). The US highway form db = V² / [30(f ± G)] with V in mph is the same physics (the 30 absorbs unit conversion and g). Perception-reaction distance is separate: 1.47 V t.

d_{b}

(ft or m)

Braking distance

V

(ft/s or m/s)

Speed (consistent units)

f

Braking friction (or a/g)

G

Grade (decimal)

g

(ft/s² or m/s²)

Gravity

Step-by-step solved example

V = 60 mph = 88 ft/s, f = 0.30, G = 0, g = 32.2 ft/s². Find db.

db = V² / [2g(f±G)] — the second term of SSD.

1. Denominator
2 g f = 2×32.2×0.30 = 19.32 ft/s².
$2 g f = 19.32$
2. Distance
db = 88² / 19.32 = 7744/19.32 = 401 ft. Check (mph form): V²/[30f] = 3600/9.0 = 400 ft.
$d_{b} = 401 ft$

Answer: 401 ft

10 practice questions

0/10 correct

1.V = 44 ft/s, f = 0.25, G = 0, g = 32.2. db is nearest

2.A downgrade G = 0.04 with f = 0.30 uses the denominator 2g times

3.Braking distance scales with

4.US customary shortcut db = V² / [30(f±G)] takes V in

5.V = 30 mph, f = 0.35, G = 0. db (US form) is nearest

6.Upgrades make db

7.Perception-reaction distance at 60 mph, t = 2.5 s, is

8.Wet pavement is handled by

9.SI: V = 20 m/s, f = 0.40, G = 0, g = 9.81. db =

10.SSD = reaction + braking. If reaction = 220 ft and db = 400 ft, SSD =