Question

A two lane urban road with one way traffic has a maximum capacity of $2400veh/hr$. Under the jam condition, the average length occupied by the vehicle is $5m$. Speed versus density relationship is linear. If traffic volume is $1000veh/hr$ then the density of the traffic (in $veh/km/2$ lane) is

• A. $23.62$
• B. $63.78$
• C. $57.62$
• D. $47.24$

Answer 1

The correct option is B $63.78$

$q_{max}=\frac{V_{sf}\times K_j}{4}$

$2400veh/hr/2lane=\frac{V_{sf}}{4}\times \left(\frac{1000}{5}\right)$

$\Rightarrow 1200=V_{sf}\times 50$

$\Rightarrow V_{sf}=24kmph$
$\because$ $q=V_{sf}(k-\frac{k^2}{k_f})$
$1000veh/hr/2lane=24(k-\frac{k^2}{200})$
$\Rightarrow 500=\frac{24}{200}(200k-k^2)$
$\Rightarrow k^2-200k+4166.67=0$
$\Rightarrow k=176.38,23.62$

Note: Higher value of $k$ is neglected because maximum density can lead to jam condition.
$\therefore k=23.62\times 2=47.24veh/km/2lane$