If a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance d40 compared to the stopping distance d25 if the driver were going at the posted speed limit of 25 mph?

Assume that the deceleration due to braking is a ft/s².

Note that

40 mph = (40/60) * 88 = 58.667 ft/s

25 mph = (25/60) * 88 = 36.667 ft/s

The final velocity is zero when the car stops, therefore

v² - 2ad = 0, or d = v² / (2a)

where

v = initial speed

a = deceleration

d = stopping distance.

The stopping distance, d₄₀, at 40 mph is

d₄₀ = 58.667² / (2a)

The stopping distance, d₂₅, at 25 mph is

d₂₅ = 36.667² / (2a)

Therefore

d₄₀/d₂₅ = 58.667² / (2a) : 36.667² / (2a)

= (58.667/36.667) ²

= 2.56

Answer:

The stopping distance at 40 mph is 2.56 times the stopping distance at 25 mph.