I think the problem would have been more naturally stated in the context of bicycles. In any case, the answer is as follows:
You are looking for an optimal velocity function $v: [0, T] \to \mathbb{R}_{\geq 0}$ satisfying some conditions. Each such function represents the strategy, "if the light is still red at time $t$, travel at speed $v(t)$; when the light turns green, coast." One of the conditions on $v$ is that you may not run the red light. In terms of the function $v$, this condition may be written as $\int_0^T v(t) \, dt \leq d$.
The quantity you wish to compute is the expected speed at which you will pass through the light after it turns green. By the givens (uniform distribution, the nature of our strategy), this expected speed is precisely the average value of $v(t)$, i.e., it is $\frac{1}{T} \int_0^T v(t) \, dt$.
Putting the last two paragraphs together, we see that the optimal expected speed is $\frac{d}{T}$. Moreover, this expected speed is achieved for any choice $v(t)$ with the property $\int_0^T v(t) \, dt = d$, i.e., for any strategy that will get you to the stoplight within time $T$.
Added in edit:
I agree with Willie Wong that maximizing the expected kinetic energy with which you pass through the light should be more physically relevant to, say, a bicyclist coasting on a shallow down-hill.
Best Answer
At Jacques' cajoling, I'm turning the comments into an answer.
The two dimensional Shubert function is just the product of the one dimensional one by itself. $f(x,y) = g(x)g(y)$ where $g(x) = \sum_{j = 1}^5 j \cos( (j+1)x + j)$ is the 1 dimensional Shubert function. Observe that the local maxima are all positive, and the local minima all negative. So the minima for $f(x,y)$ occur at points $\{(x,y) : g'(x)= g'(y) = 0, f(x,y) < 0\}$. In other words, the minima of $f$ occurs at points where a maximum of $g$ is multiplied against a minimum.
Notice that there are 3 global max/min each of $g$ in the interval (-10,10), and 19 max and 20 min overall. This produces the 760 total local min of $f$ with 18 of them being global. (760 = 2 * 19 * 20, 18 = 2 * 3 * 3)
To find the extrema of the 1-d Shubert function, you evaluate its first derivative, and find that it can be simplified to a degree 6 polynomial in $\sin(x)$ and $\cos(x)$ by using the angle addition formulae. I have not evaluated the computations myself, so cannot tell you whether the expression has a closed-form algebraic solution.