Question

The free flow speed of a highway is 100 km/h and its capacity is 4000 vehicle/h. Assume speed density relation is linear. For a traffic volume of 2000 vehicle/h, choose all the possible speeds (in km/h) from the options given below (round off to two decimal places).

• A. 85.36
• B. 65.20
• C. 14.64
• D. 7.22

Hint:

In problems involving speed-density relations, remember that the traffic flow equation relates the speed and density. Solve for unknown speeds using the given equations and check against the options for possible values.

Answer 1

The Correct Option is A, C

Given: Free mean speed, \( V_f = 100 \, {kmph} \)
Capacity, \( q_{{max}} = 4000 \, {veh/hr} \)
Possible speed, \( V_1, V_2 = ? \)
Speed-density relation is linear. The equation for speed-density relation is: \[ q_{{max}} = \frac{1}{4} k_j V_f \] Substituting known values: \[ 4000 = \frac{1}{4} \times 100 k_j \] This simplifies to: \[ k_j = 160 \, {veh/km} \] Now, using the formula for traffic flow, \( q = v k \), we get: \[ q = v_f \left( k - \frac{k^2}{k_j} \right) \] Substituting the values: \[ 2000 = 100 \left( k - \frac{k^2}{160} \right) \] On solving, we get: \[ k_1 = 23.431 \, {veh/km}, \quad k_2 = 136.568 \, {veh/km} \] Now, the velocity of traffic flow at \( k_1 = 23.431 \, {veh/km} \) is: \[ V_1(k_1 = 23.431) = 100 \left( 1 - \frac{23.431}{160} \right) = 85.355 \, {km/hr} \] The velocity of traffic flow at \( k_2 = 136.568 \, {veh/km} \) is: \[ V_2(k_2 = 136.568) = 100 \left( 1 - \frac{136.568}{160} \right) = 14.645 \, {km/hr} \] Thus, the correct options are (a) and (c).