As a city planner, you receive complaints from local residents about the safety of nearby roads and streets. One complaint concerns a stop sign at the corner of Pine Street and 1st Street. Residents complain that the speed limit in the area (55 mph) is too high to allow vehicles to stop in time. Under normal conditions this is not a problem, but when fog rolls in visibility can reduce to only 155 feet. Since fog is a common occurrence in this region, you decide to investigate. The state highway department states that the effective coefficient of friction between a rolling wheel and asphalt ranges between 0.536 and 0.599, whereas the effective coefficient of friction between a skidding (locked) wheel and asphalt ranges between 0.350 and 0.480. Vehicles of all types travel on the road, from small VW bugs weighing 1150 lb to large trucks weighing 9360 lb. Considering that some drivers will brake properly when slowing down and others will skid to stop,
a) calculate the miminim and maximum braking distance needed to ensure that all vehicles traveling at the posted speed limit can stop before reaching the intersection.
b) Given that the goal is to allow all vehicles to come safely to a stop before reaching the intersection, calculate the maximum desired speed limit. (Round your answer to the nearest whole number.)

A) i)The maximum braking distance needed = 290.5 ft

    ii)The minimum braking distance needed = 169.74 ft

B) The maximum desired speed limit ( ) = 41 mph  

A) Determine the minimum and maximum braking needed

Applying Newton's second law

along x-axis : Fr = ma ---- ( 1 )

along y-axis : N = mg ----- ( 2 )

force of friction ( fr ) = μ N ----- ( 3 )

Equating equations ( 1 ), ( 2 ) and ( 3 )

 μ*mg = ma

 ∴  a = μ g  ---- ( 4 )    where g = 32 ft/s²

next step : calculate the acceleration value for each coefficient ( μ )

for μ  = 0.350    a₁ = 11.20 ft/s²  ( minimum value )

for μ  = 0.599    a₂ = 19.168 ft/s²  ( maximum value )

V₀ = 55 mph*(5280 ft / 1 mile)*( 1h / 3600s)  = 80.667 ft / s²

i) Maximum braking distance

x₁ = V₀² / 2 * a₁  

∴ The maximum braking distance = ( 80.667 )² / ( 2 *  11.200 ) = 290.5 ft

ii) Minimum braking distance

x₂ = V₀² / 2 * a₂  

∴ x₂ = ( 80.667 )² / ( 2 *  19.168 )  =  169.74ft

B) Calculate The maximum desired speed limit ( )

V₀ = √ 2 * a* x

where ; a = 11.2 ft/s²  and  19.168 ft/s²

∴ V₀ = 58.92 ft/s  and  77.08 feet / s

Therefore  the maximum desired speed limit  = 58.92 ft/s = 41 mph

Hence we can conclude that the, The maximum braking distance needed = 290.5 ft ,The minimum braking distance needed = 169.74 ft, The maximum desired speed limit ( ) = 41 mph.

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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

electrical power

true tan speed = ang vel x radius