Can you help me with a difficult physics problem involving constant acceleration

First, let it be said that in my class only two equations are allowed to be used at this point in time. They are #1: Final Velocity = Initial Velocity + Acceleration x Time and #2: Distance = Average Velocity x Time

We are supposed to solve all problems by using and manipulating these two formulas. So...the problem is:

"An object moving with uniform acceleration has a velocity of 12.0 cm/s in the positive x direction when its x coordinate is 3 cm. If its x coordinate 2.00 seconds later is -5.00 cm, what is its acceleration?"

How can this possibly be solved using the two equations above? There are too many unknown variables. But the teacher insists it can be done. Calculus 1 is the math prerequisite to the class, if that makes a difference. Please help me understand this!

I think you have to make use of eq. for average velocity under constant acceleration as;
Average Velocity = (Final Velocity + Initial Velocity)/2

Using that in eq. 2 will allow you to find the Final Velocity, which you can use to then find the acceleration.

I know that's bringing in a third eq. ,but it comes from the definition of the average of any function that increases proportionally on an interval; from calculus;

The average value of any function f(x) on an interval L = x2 - x1 is
f(avg) = (1/L)INT[f(x)dx] from x1 to x2

If f(x) = ax (proportional to x) then the integral is;
f(avg) = (a/2L)(x2^2 - x1^2) = (a/2)(x2 + x1) = (f2 + f1)/2

Without that relationship for average you can't solve your problem.

Another key, I realized, is that equation #2 actually involves displacement, not distance. Knowing that, I could figure the average, solve for the Final Velocity, go to equation #1 and find the acceleration, which in this case turns out to be -16 cm/s^2. Nice tricky 1st day of class problem.

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