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Traffic flow equation for a section of road is u = 80 - 0.7 K where ‘u’ is the speed in kmph and ‘K’ is the density in vpkm (vehicles per km). The maximum expected flow is
- a)4572 vph
- b)2286 vph
- c)1143 vph
- d)572 vph
Correct answer is option 'B'. Can you explain this answer?
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Traffic flow equation for a section of road is u = 80 - 0.7 K where &l...
Traffic flow equation:
u = 80 - 0.7k
Since flow, q = u · k = 80 k - 0.7 k2
u = 80 - 0.7k
Since flow, q = u · k = 80 k - 0.7 k2
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Traffic flow equation for a section of road is u = 80 - 0.7 K where &l...
Understanding the Traffic Flow Equation
The traffic flow equation given is:
u = 80 - 0.7K
where:
- u = speed in km/h
- K = density in vehicles per km (vpkm)
Finding the Flow (Q)
Flow can be expressed as:
Q = u * K
Substituting the equation for u:
Q = (80 - 0.7K) * K
This simplifies to:
Q = 80K - 0.7K²
Maximizing the Flow
To find the maximum flow, we need to take the derivative of Q with respect to K and set it to zero:
dQ/dK = 80 - 1.4K = 0
Solving for K gives:
1.4K = 80
K = 80 / 1.4
K ≈ 57.14 vpkm
Calculating Maximum Flow
Now, substituting K back into the flow equation:
Q = 80K - 0.7K²
Q = 80(57.14) - 0.7(57.14)²
Calculating each part:
- 80 * 57.14 ≈ 4571.2
- 0.7 * (57.14)² ≈ 0.7 * 3265.62 ≈ 2286
Thus, the maximum expected flow is:
Q ≈ 4571.2 - 2286 ≈ 2285.2 vph
Conclusion
The maximum expected flow is approximately 2286 vph, confirming that the correct answer is option 'B'.
The traffic flow equation given is:
u = 80 - 0.7K
where:
- u = speed in km/h
- K = density in vehicles per km (vpkm)
Finding the Flow (Q)
Flow can be expressed as:
Q = u * K
Substituting the equation for u:
Q = (80 - 0.7K) * K
This simplifies to:
Q = 80K - 0.7K²
Maximizing the Flow
To find the maximum flow, we need to take the derivative of Q with respect to K and set it to zero:
dQ/dK = 80 - 1.4K = 0
Solving for K gives:
1.4K = 80
K = 80 / 1.4
K ≈ 57.14 vpkm
Calculating Maximum Flow
Now, substituting K back into the flow equation:
Q = 80K - 0.7K²
Q = 80(57.14) - 0.7(57.14)²
Calculating each part:
- 80 * 57.14 ≈ 4571.2
- 0.7 * (57.14)² ≈ 0.7 * 3265.62 ≈ 2286
Thus, the maximum expected flow is:
Q ≈ 4571.2 - 2286 ≈ 2285.2 vph
Conclusion
The maximum expected flow is approximately 2286 vph, confirming that the correct answer is option 'B'.
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