Answer:
a) x₁ = 290.50 feet
, x₂ = 169.74 feet
, b) v_max= 41 mph
Explanation:
For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles
X Axis fr = m a
Y Axis N-W = 0
N = W = mg
The force of friction has the expression
fr = μ N
We replace
μ mg = ma
a = μ g
g = 32 feet / s²
Let's calculate the acceleration for each coefficient and friction
μ a (feet / s2)
0.599 19.168
0.536 17,152
0.480 15.360
0.350 11.200
These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)
v² = v₀² - 2 a x
When the speed stops it is zero
x₁ = v₀² / 2 a₁
Let's reduce speed
v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²
Let's calculate the maximum braking distance
x₁ = 80.667² / (2 11.2)
x₁ = 290.50 feet
The minimum braking distance
x₂ = 80.667² / (2 19.168)
x₂ = 169.74 feet
b) maximum speed to stop at distance x = 155 feet
0 = v₀² - 2 a x
v₀ = √2 a x
We calculate the speed for the two accelerations
v₀₁ = √ (2 11.2 155)
v₀₁ = 58.92 feet / s
v₀₂ = √ (2 19.168 155)
v₀₂ = 77.08 feet / s
To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph
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