Let $v_0$ be the constant speed of the traffic when moving (during the green signal). When the signal is red, the speed is 0.

• Let the duration of the green signal be $T_G = T$.
• Let the duration of the red signal be $T_R = T$ (since they are initially equal).
• Total cycle time: $t_1 = T_G + T_R = 2T$.

The distance covered in one cycle is simply the speed multiplied by the moving time:

$$ d_1 = v_0 \times T_G = v_0 T $$

The average speed $v_1$ is total distance divided by total time:

$$ v_1 = \frac{d_1}{t_1} = \frac{v_0 T}{2T} = \frac{v_0}{2} $$

From the problem, we know $v_1 = 1.5 \, \text{m/s}$, which implies the cruising speed is $v_0 = 3.0 \, \text{m/s}$.

The duration of the green signal is increased by a factor of $\eta = 2$, while the red signal duration remains unchanged.

• New Green duration: $T’_G = \eta T = 2T$.
• New Red duration: $T’_R = T$.
• New Total cycle time: $t_2 = T’_G + T’_R = 2T + T = 3T$.

The new distance covered in one cycle:

$$ d_2 = v_0 \times T’_G = v_0 (2T) $$

The new average speed $v_2$ is:

$$ v_2 = \frac{d_2}{t_2} = \frac{v_0 (2T)}{3T} = \frac{2}{3} v_0 $$

Substituting $v_0 = 2v_1$ from the first step:

$$ v_2 = \frac{2}{3} (2v_1) = \frac{4}{3} v_1 $$