There is a slight mistake in your analysis towards the beginning. You assumed that the speed is equal to the total acceleration time the time. However the radial acceleration has no effect on the magnitude of the velocity in circular motion.
Therefore we can write,
$$
v(t) = a_t t
$$
Now lets assume your target velocity is $v_{max}$ then we know ahead of time that the maximum radial acceleration will be,
$$ a_{r_{max}} = \frac{v_{max}^2}{R}$$
Since we also know the maximum total acceleration, $a_{max}$ we can easily compute the maximum allowed tangential acceleration.
$$(a_{max})^2 = (a_r)^2 + (a_t)^2$$
$$a_t = \sqrt{(a_{max})^2-(a_{r_{max}})^2}$$
$$a_t = \sqrt{(a_{max})^2-\left(\frac{V_{max}^2}{R}\right)^2}$$
We know that the quantity inside the square root is strictly positive because the radial acceleration never exceeds the total acceleration.
The speed on the first leg of the process is then,
$$
v_I(t) = a_t t ; t \leq V_{max}/a_t
$$
The angular distance for this step is given by
$$
\Theta_I = \int_0^{V_{max}/a_t} v_I(t)/R dt = \frac{V_{max}^2}{2Ra_t}
$$
The angular distance on the third step of the process should be the same since the time for decelleration and the average speed will remain the same.
$$
\Theta_{III} = \Theta_{I} = \frac{V_{max}^2}{2Ra_t}
$$
On the second leg of the journey we travel at a constant speed. The amount of time this takes depends on how much of the arc is left. Let $\Delta \Theta$ be the total angle traveled then we have,
$$\Delta \Theta = \Theta_I + \Theta_{II} + \Theta_{III}$$
$$\Theta_{II} = \Delta \Theta - (\Theta_I + \Theta_{III})$$
$$\Theta_{II} = \Delta \Theta -\frac{V_{max}^2}{Ra_t}$$
The time to traverse this arc is then,
$$ \Delta T_{II} = \frac{\Theta_{II} R}{ V_{max}}$$
Thet total time is then the time elapsed for each of the three journeys,
$$ T_{total} = \left( \frac{V_{max}}{a_t}\right) + \left( \frac{\Theta_{II} R}{ V_{max}}\right) + \left( \frac{V_{max}}{a_t} \right)$$
$$T_{total} = \left( \frac{2V_{max}}{a_t}\right) + \left( \frac{\Theta_{II} R}{ V_{max}}\right) $$
Everything on the right side of this equation is known so all that is left is some simple substitutions. This is all assuming that the tangential acceleration is constant, if you are able to vary the tangential acceleration then a more general solution using elliptic integrals is possible.
Best Answer
From the area of the trapezium you have $$(t_3+4)\frac V2=1000$$
From the suvat formula $v=u+at$ you also have $$V=1000t_1=500t_2$$
And, of course, $$t_1+t_2+t_3=4\implies t_3=4-\frac{3V}{1000}$$
This leads to the quadratic equation $$\left(8-\frac{3V}{1000}\right)\frac V2=1000$$
Solve this and you get $$V=279.24$$ as well as another, spurious, solution. The other results then follow.