Stopping Sight Distance While descending a -7% grade at a speed of 90 km/h, George notices a large object in the roadway ahead of him. Without thinking about any alternatives, George stabs his brakes and begins to slow down. Assuming that George is so paralyzed with fear that he won't engage in an avoidance maneuver, calculate the minimum distance at which George must have seen the object in order to avoid colliding with it. You can assume that the roadway surface is concrete and that the surface is wet.  You can also assume that George has a brake reaction time of 0.9 seconds because he is always alert on this stretch of the road. [Solution Shown Below]                       Solution First, we need to calculate the distance that George traveled during his brake reaction time. This is done using the equation D = VT from physics. Since George's brake reaction time was 0.9 seconds and his velocity was 25 m/sec (90 km/h), the distance he traveled during his brake reaction time was 22.5 meters.  Second, we need to calculate the distance George will travel while braking. This is done using the equation shown below. d = V2/(2g(f + G)) Where: V = George's Velocity, 25 m/sec (90 km/h) g = Acceleration due to gravity, 9.81 m/sec2 f = Coefficient of friction, 0.29 ( we'll use the value for 96 km/h (60 mph) just to be conservative) G = The grade of the road, -0.07 (-7%) Solving the equation yields a distance of 145 meters. Summing the distance traveled while braking and the distance traveled during the brake reaction time yields a total stopping sight distance of 168 meters, which is about 15.2 meters short of two football fields. George needed to be about 168 meters away from the object at the instant he first saw it in order to avoid a collision.