Motion problem with Acceleration and distance

[risk314]

Homework Statement

A cabdriver picks up a woman. The driver speeds up at a constant acceleration until he reaches the speed limit and then instantaneously begins decelerating at a constant rate until he reaches a velocity of 0. The magnitude of the deceleration is 3 times that of the acceleration. The distance from the point at which the woman is picked up to the point at which she is dropped off is 3.2km.

Homework Equations

The 4 motion equations are relevant but I'm not sure how.

The Attempt at a Solution

I tried plugging in the variables into "V²=V_i²+2ad" by setting it equal to itself with 1 of them having just a and d and one having 3a and d₁ but I ended up just getting to different d's on separate sides of an equation.

 

[Delphi51]

I can't seem to solve it, but perhaps I can offer some notation to get you started.
Let t be the time and "a" the acceleration for the first part.
Then -3a is the acceleration for the second part.
We have that the velocity gained in the first part is lost in the second part so
at - 3aT = 0 where T is the time for the second part, and T = t/3.
The next step would be to write that the distance in the first part plus the distance in the second part = 3200 m. That is as far as I get - there are two unknowns (a and t) and no apparent way to find them.

What is the question?