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inna [77]
4 years ago
11

A minor road intersects a major 4-lane divided road with a design speed of 55mph and a median width of 8 feet. The intersection

is controlled with a stop sign on the minor road. If the design vehicle is a passenger car, determine the minimum sight distance required on the major road that will allow a stopped vehicle on the minor road to safely turn left if the approach grade on the minor road is 3%.

Engineering
1 answer:
Vedmedyk [2.9K]4 years ago
8 0

Answer:

stopping sight diatnce is =158 meters

Explanation:

The Design Speed =55 MPH(Miles per Hour)

1 miles per hour =0.447 meter pe second

then 55MPH=55*0.447=24.58 meter per second

Consider recation time =2.5 second

Consider Co efficinet friction is 0.35

As per the Stopping sight distance based on the Breaking distance+ Lag distance

SSD=Vt+(V^2/(2gf))

SSD=24.58+(24.58^2/(2*9.81*0.35))

The total stopping sight distance is =150 meters

if we consider the Approaching garde on the minor road is 3%

SSD=Vt+(V^2/(2g(f-n/100))

SSD=24.58+(24.58^2/(2*9.81*(0.35-3/100))

then stopping sight diatnce is =158 meters

See attachment for workings

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A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude
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Answer:

u = 260.22m/s

S_{max} = 1141.07ft

Explanation:

Given

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S_{16.5} = 0ft -- Altitude after 16.5 seconds

a = -g = -32.2ft/s^2 --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

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To do this, we make use of:

S = ut + \frac{1}{2}at^2

The final altitude after 16.5 seconds is represented as:

S_{16.5} = S_0 + ut + \frac{1}{2}at^2

Substitute the following values:

S_0 = 89.6ft       S_{16.5} = 0ft     a = -g = -32.2ft/s^2    and t = 16.5

So, we have:

0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2

0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45

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Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

v=u  + at

At the maximum height:

v =0 --- The final velocity

u = 260.22m/s

a = -g = -32.2ft/s^2

So, we have:

0 = 260.22 - 32.2t

Collect Like Terms

32.2t = 260.22

Make t the subject

t = \frac{260.22}{ 32.2}

t = 8.08s

The maximum height is then calculated as:

S_{max} = S_0 + ut + \frac{1}{2}at^2

This gives:

S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2

S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22

S_{max} = 89.6 + 260.22 * 8.08 - 1051.11

S_{max} = 1141.0676

S_{max} = 1141.07ft

Hence, the maximum height is 1141.07ft

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