On a level test track, a car with antilock brakes and 90% braking efficiency is determined to have a theoretical stopping distance (ignoring aerodynamic resistance) of 408 ft (after the brakes are applied) from 100 mi/h. The car is rear-wheel drive with a 110-inch wheelbase, weighs 3200 lb, and has a 50/50 weight distribution (front and back), a center of gravity that is 22 inches above the road surface, an engine that generates 300 ft-lb of torque, and overall gear reduction of 8.5 to 1 (in first gear), a wheel radius of 15 inches and a driveline efficiency of 95%. What is the maximum acceleration from the rest of this car on this test track

To solve this problem let's start by finding the braking acceleration using kinematics, where the distance is x = 408 ft, the initial velocity vo = 100 mi / h and the final velocity is zero v = 0

v² = v₀² - 2 a x

0 = v₀² - 2ax

a = [tex]\frac{v_o^2}{2x}[/tex]

Let's start by reducing the magnitudes to ft / s

v₀ = 100 mi / h (5280 foot / 1 mile) (1h / 3600 s) = 146.666 ft / s

let's calculate

a = [tex]\frac{146.66^2}{2 \ 408}[/tex]

a = 26.36 ft / s²

Let's call this acceleration a_effective, this acceleration is in the opposite direction to the speed of the vehicle.

Let's use a rule of three (direct proportions) to find the acceleration applied by the brake system (a1) which has an efficiency of 95%. or 0.95

a₁ = [tex]\frac{a_e}{0.95}[/tex]

Let's use another direct proportion rule If the acceleration of the brake system (a₁) for an applied acceleration (a) with an efficiency of 0.90

a = [tex]\frac{a_1}{0.90}[/tex]

we substitute

a = [tex]\frac{a_e}{0.95 \ 0.90}[/tex]

let's calculate

a = [tex]\frac{26.36}{ 0.95 \ 0.90}[/tex]

a = 30.832 ft/s²

This is the maximum relationship that the vehicle can have for when it brakes to stop at the given distance