Chapter 5, Problem 18P

To determine

(i)

The long- time average number of injury crashes per year at the site.

Answer to Problem 18P

  $\underset{\_}{n_i=14.724Crashes/yr.}$

Explanation of Solution

Given:

Following information has been given to us: Length of road segment $=7585ft.$

AADT $=5250veh/day.$

Total number of injury crashes for 10 years $=12.$

Total number of fatal crashes for 10 years $=3.$

Total number of property crashes for 10 years $= 48.$

Assume $k=2.51.$

The applicable safety performance functions for injury, fatal, and property damage only

crashes are:

  $S{P}_{inj}=1.602\times {10}^{-6}\left(lengthinfeet\right){\left(AADT\right)}^{0.491}$

  $S{P}_{PDO}=3.811\times {10}^{-6}\left(lengthinfeet\right){\left(AADT\right)}^{0.491}$

Calculation:

We have the following formula for the number of fatal crashes:

  $n_i=\frac{\text{number}\text{of}\text{fatal}\text{crashes}}{S{P}_{inj}}$

Where, $S{P}_{inj}=1.602\times {10}^{-6}\left(lengthinfeet\right){\left(AADT\right)}^{0.491}$

  $\begin{array}{l}S{P}_{inj}=1.602\times {10}^{-6}\left(7585\right){\left(5250\right)}^{0.491}\\ S{P}_{inj}=0.815\end{array}$

Substituting the values, we have

  $\begin{array}{l}{n}_i=\frac{\text{12}}{0.815}\\ n_i=14.724.\end{array}$

Conclusion:

Therefore, the long- time average number of injury crashes per year at the site is $n_i=14.724$ Crashes/ yr.

To determine

(ii)

The long-time average number of fatal crashes per year at the site.

Answer to Problem 18P

  $\underset{\_}{n_f=3.681\text{Crashes/ year}}.$

Explanation of Solution

Given:

Following information has been given to us:

Length of road segment $=7585ft.$

AADT $=5250veh/day.$

Total number of injury crashes for 10 years $=12.$

Total number of fatal crashes for 10 years $=3.$

Total number of property crashes for 10 years $= 48.$

Assume $k=2.51.$

The applicable safety performance functions for injury, fatal, and property damage only

crashes are:

  $S{P}_{inj}=1.602\times {10}^{-6}\left(lengthinfeet\right){\left(AADT\right)}^{0.491}$

  $S{P}_{PDO}=3.811\times {10}^{-6}\left(lengthinfeet\right){\left(AADT\right)}^{0.491}$

Calculation:

We have the following formula for the number of fatal crashes:

  $n_f=\frac{\text{number}\text{of}\text{fatal}\text{crashes}}{S{P}_{inj}}$

Where, $S{P}_{inj}=1.602\times {10}^{-6}\left(lengthinfeet\right){\left(AADT\right)}^{0.491}$

  $\begin{array}{l}S{P}_{inj}=1.602\times {10}^{-6}\left(7585\right){\left(5250\right)}^{0.491}\\ S{P}_{inj}=0.815.\end{array}$

Substituting the values, we have

  $\begin{array}{l}{n}_f=\frac{3}{0.815}\\ n_f=3.681\end{array}$

  $n_f=3.681\text{Crashes/ year}\text{.}$

Conclusion:

Therefore, the number of fatal crashes per year at the site is $3.681\text{Crashes/ year}\text{.}$

To determine

(iii)

The long-time average number of property damage only crashes per year at the site.

Answer to Problem 18P

  $\underset{\_}{n_P=24.755\text{Crashes/ year}\text{.}}$

Explanation of Solution

Given:

Following information has been given to us:

Length of road segment $=7585ft.$

AADT $=5250veh/day.$

Total number of injury crashes for 10 years $=12.$

Total number of fatal crashes for 10 years $=3.$

Total number of property crashes for 10 years $= 48.$

Assume $k=2.51.$

The applicable safety performance functions for injury, fatal, and property damage only

crashes are:

  $S{P}_{inj}=1.602\times {10}^{-6}\left(lengthinfeet\right){\left(AADT\right)}^{0.491}$

  $S{P}_{PDO}=3.811\times {10}^{-6}\left(lengthinfeet\right){\left(AADT\right)}^{0.491}$

Calculation:

We have the following formula for the long-time average number of property damage only crashes per year at the site

  $n_P=\frac{\text{number}\text{of}\text{property}\text{damage crashes}}{S{P}_{PDO}}$

Where, ${\text{SP}}_{PDO}=3.811\times {10}^{-6}\left(lengthinfeet\right)\times \left(AADT\right){}^{0.491}$

  $\begin{array}{l}{\text{SP}}_{PDO}=3.811\times {10}^{-6}\left(7585\right)\times {\left(5250\right)}^{0.491}\\ {\text{SP}}_{PDO}=1.939\\ {\text{SP}}_{PDO}=1.939\text{Crashes}\text{/year}\text{.}\end{array}$

Substituting the values, we have

  $\begin{array}{l}{n}_P=\frac{48}{1.939}\\ n_P=24.755\\ n_P=24.755\text{Crashes/ year}\text{.}\end{array}$

Conclusion:

Therefore,the long-time average number of property damage only crashes per year at the site $n_P=24.755\text{Crashes/ year}\text{.}$