Question

A car is traveling at 70 mi/h on a level section of road with good, wet pavement. Its antilock braking system (ABS) only starts to work after the brakes have been locked for 100 ft. If the driver holds the brake pedal down completely, immediately locking the wheels, and keeps the pedal down during the entre process, how many feet will it take the car to stop from the point of initial brake application? (The braking efficiency is 80% with the ABS not working and 100% with the ABS working. Use theoretical stopping distance and ignore air resistance. Let frl = 0.02 when the brakes are locked, but compute the frl once the ABS becomes active.)

Answer 1

Answer: The theoretical stopping distance is 1.245 miles (6572.193 ft).

Explanation:

First, the physical model is created by using Principle of Energy Conservation and Work-Energy Theorem. It is assumed that surface is horizontal, so there are no changes associated with potential energy. The car has an initial kinetic energy, which is completely dissipated by braking.

$K_{1} = \Delta W_{loss}$

$\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot (\mu_{2} \cdot \Delta s_{2}+\mu_{2'} \cdot \Delta s_{2'})$

The distance is isolated from previous equation:

$\Delta s_{2'} = \frac{1}{\mu_{2'}}\cdot (\frac{1}{2}\cdot \frac{v^{2}}{g} - \mu_{2} \cdot \Delta s_{2})$

By replacing variables, the distance is calculated herein:

$\Delta s_{2'} = \frac{1}{\frac{0.02}{0.80}}\cdot (\frac{1}{2}\cdot \frac{(102.667\frac{ft}{s} )^2}{32.174 \frac{ft}{s^{2}}} - (0.02) \cdot (100 ft))$

$\Delta s_{2'} = 6472.193 ft\\\Delta s_{2'} = 1.226 mi$

The theoretical stopping distance is:

$\Delta s = \Delta s_{2} + \Delta s_{2'}\\\Delta s = 6572.193 ft\\\Delta s = 1.245 mi$