If a child ran into the road 65 to 70 feet ahead of your vehicle, what is the highest speed from which you could stop with good brakes before hitting him?
Let v = the highest speed of the vehicle. Let d = the stopping distance.
The formula for stopping distance is d = v²/(2μg) where μ = 0.8, the static coefficient of friction for good brakes (normal conditions) g = 32.2 ft/s², acceleration due to gravity.
In terms of v, v = √(2μgd)
Note that 88 ft/s = 60 mph.
Consider d = 65 ft. v = √2*0.8*32.2*65) = 57.87 ft/s or v = 57.87*(60/88) = 39.5 mph
Consider d = 70 ft. v = √(2*0.8*32.2*70) = 60.05 ft/s or v = 60.05*(60/88) = 40.95 mph
The lower of these two speeds should be the highest allowable speed in order to avoid hitting the child.
Answer: The highest speed is 39.5 mph or 57.9 ft/s.
