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Engineering
19.06.2020 22:38
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A road has a crest curve, where the PVI station is a 71 35. The road transitions

A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest point of the curve is at station 74 10. What are the PVC and PVT stations
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2022rickskyye
2022rickskyye
4,6(62 marks)

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

y = ax^{2} +bx+c\\dy/dx=2ax+b\\

At highest point of the curve dy/dx=o

2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275

-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\

Station PVT

Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)

snowprincess99447
snowprincess99447
4,9(52 marks)

y +6 = -(x +2)

Step-by-step explanation:

Given

Perpendicular to: y = x - 7

Passes through (-2,-6)

Required

Find the equation

An equation has the form:

y = mx + b

Where

m = slope

In y = x - 7

m = 1

Since the required equation is perpendicular to y = x - 7, then the relationship between their slopes is:

m_2 = -\frac{1}{m}

Substitute 1 for m

m_2 = -\frac{1}{1}

m_2 = -1

The equation is then calculated using:

y -y_1 = m(x - x_1)

Where

m_2 = -1

(x_1,y_1) = (-2,-6)

So, we have:

y - (-6) = -1(x - (-2))

y +6 = -1(x +2)

y +6 = -(x +2)

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