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A driver is traveling at 52 mi/h on a wet road. an object is spotted on the road 415 ft ahead and the driver is able to come to a stop just before hitting the object. assuming standard perception/ reaction time and practical stopping distance, determine the grade of the road.

Respuesta :

Answer:

G = 0.424

Explanation:

Ds = ( 0.278tr * V ) + (0.278 * V²)/ ( 19.6* ( f ± G))

Where Ds = stopping sight distance = 415miles = 126.5m

G = absolute grade road

V = velocity of vehicle = 52miles/hr

f = friction = 0 because the road is wet

tr = standard perception / reaction time = 2.5s

So therefore:

Substituting to get G

We have

2479.4G = 705.6G + 751.72

1773.8G = 751.72

G = 751.72/1773.8

G = 0.424

Lanuel

Based on the calculations, the grade of the road is equal to 2.43 %.

Given the following data:

  • Velocity = 52 mi/h to km/h = 83.6859 km/h.

Assumed data:

  • Acceleration = 3.5 [tex]m/s^2[/tex]
  • Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]
  • Standard perception/reaction time = 2.5 seconds
  • Stopping distance = 135 meters.

The formula for stopping distance.

Mathematically, the stopping distance for an inclined surface with a coefficient of friction is given by this formula:

[tex]SD = 0.278Vt + \frac{V^2}{254} (f \pm 0.01G)[/tex]

Where:

  • V is the velocity.
  • t is the perception/reaction time
  • G is the grade of a surface.
  • f is the coefficient of friction.

Substituting the given parameters into the formula, we have;

[tex]135 = 0.278 \times 83.6859 \times 2.5 + [\frac{83.69^2}{254} \times (\frac{3.5}{9.8} + 0.01G)]\\\\135 = 58.14 + [27.58 \times (0.36 + 0.01G)]\\\\135 = 58.14 +9.93+0.2758G\\\\0.2758G=135 - 58.14 -9.93\\\\0.2757G=66.9923\\\\G=\frac{66.93}{0.2757}[/tex]

G = 242.76 ≈ 243

G = 2.43 %

Read more on standard perception/reaction time here: https://brainly.com/question/13218365

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