Answer
$t_r+\frac{v_p}{a}+\frac{D-d-\frac{v_p^2}{2a}}{v_p}$
Work Step by Step
Let $t_1$ be the time to accelerate to a speed of $v_p$
Let $t_2$ be the time to reach a distance $d$ from intersection 2 when the speed is $v_p$
The time delay should be $t_r+t_1+t_2$
We can find $t_1$:
$t_1 = \frac{v_p}{a}$
We can find the distance traveled during the acceleration period:
$x_1 = \frac{1}{2}at_1^2$
$x_1 = \frac{1}{2}a(\frac{v_p}{a})^2$
$x_1 = \frac{v_p^2}{2a}$
We can find the remaining distance:
$x_2 = D-d-x_1$
$x_2 = D-d-\frac{v_p^2}{2a}$
We can find $t_2$:
$t_2 = \frac{x_2}{v_p}$
$t_2 = \frac{D-d-\frac{v_p^2}{2a}}{v_p}$
We can find the total time delay between the green lights:
$t_r+t_1+t_2 = t_r+\frac{v_p}{a}+\frac{D-d-\frac{v_p^2}{2a}}{v_p}$
