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The distance it takes a truck to stop can be modeled by the function;
d(u) = 2.15u²/64.4f
Where
d = stopping distance in feet
u = initial velocity in miles per hour
f = a constant related to friction
When the truck's initial velocity on dry pavement is 40 mph, its stopping distance is 138 ft.
Determine the value of f.
Answer:
The friction constant is approximately 0.3871
Step-by-step explanation:
Given
Stopping distance, d(u) = 138ft
Initial Velocity, u = 40mph
Required
Friction constant, f.
To get the value of f, we need to simply substitute 40 for u and 138 for d(u) in the above expression.
In other words;
d(u) = 2.15u²/64.4f becomes
138 = 2.15 * 40²/64.4f
138 = 2.15 * 1600/64.4f
138 = 3440/64.4f
Multiply both sides by 64.4f
138 * 64.4f = 64.4f * 3440/64.4f
8887.2f = 3440
Divide both sides by 8887.2
8887.2f/8887.2 = 3440/8887.2
f = 3440/8887.2
f = 0.387073544
f = 0.3871 (Approximated)
Hence, the friction constant is approximately 0.3871
