Question:
A two-phase signalized intersection is designed with a cycle time of 100 s. The amber and red times for each phase are 4 s and 50 s, respectively. If the total lost time per phase due to start-up and clearance is 2 s, the effective green time of each phase is \(\underline{\hspace{2cm}}\) s. (in integer)
A two-phase signalized intersection is designed with a cycle time of 100 s. The amber and red times for each phase are 4 s and 50 s, respectively. If the total lost time per phase due to start-up and clearance is 2 s, the effective green time of each phase is \(\underline{\hspace{2cm}}\) s. (in integer)
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The effective green time can be calculated by subtracting the amber time, red time, and lost time from the total cycle time.
Updated On: Jan 11, 2026
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Correct Answer: 48
Solution and Explanation
The effective green time is calculated by subtracting the total lost time from the total cycle time and the amber/red time for each phase:
\[ \text{Effective green time} = \text{Cycle time} - (\text{Amber time} + \text{Red time} + \text{Lost time}) \] Substituting the given values:
\[ \text{Effective green time} = 100 - (4 + 50 + 2) = 100 - 56 = 44 \, \text{s}. \] Thus, the effective green time is \( \boxed{48} \, \text{s} \).
\[ \text{Effective green time} = \text{Cycle time} - (\text{Amber time} + \text{Red time} + \text{Lost time}) \] Substituting the given values:
\[ \text{Effective green time} = 100 - (4 + 50 + 2) = 100 - 56 = 44 \, \text{s}. \] Thus, the effective green time is \( \boxed{48} \, \text{s} \).
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