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]

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****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%)