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

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

Wish i knew this to !