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The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the road if (a) the road is level and (b) the road has a maximum grade of 4%? Assume the perception-reaction time = 2.5 sec

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danialamin

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

[tex]SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}[/tex]

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              [tex]SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft[/tex]

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

[tex]SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}[/tex]

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              [tex]SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft[/tex]

So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.

For downgrade of 4%, Substituting values

                              [tex]SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft[/tex]

So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

okpalawalter8

Answer:

  • The minimum sight distance = [tex]161.05m[/tex]

Explanation:

Given that

Speed, [tex]V = 60mph = 96.552Km/h = 26.82m/s[/tex]

Accending gradient = [tex]4\%[/tex]

Perception reaction time = [tex]2.5sec[/tex]

For [tex]V = 80Km/h[/tex]

Coefficient of longitudenal friction [tex]f = 0.35[/tex]

Therefore,

Stopping sight distance,

[tex]SSD = Vt + \frac{V^2}{2g(f+0.01n)}\\\\SSD = 26.82*2.5 + \frac{26.82^2}{2*9.81(0.35+0.01*4)}\\\\SSD = 161.05m[/tex]

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https://brainly.com/question/14618740