I'm not sure if Advanced Algebra is the correct forum for this, but if not maybe it can be moved elsewhere?

Anyways, I am taking this class where we want to solve coefficients for a cubic polynomial. This handout reads:

Suppose we want to find the motion trajectory for a path with only two knot points - the start and goal, which satisfier the time constraint that the tie duration of motion is T, and in addition, satisfier the velocity constraint that the omotion starts from zero velocity and ends with zero velocity. Then, for each variable q_{i}, its trajectory as a function of time *t* should satisfy the following 4 constraints:

- q
_{i}(0) = s_{i} - q
_{i}'(0) = 0 - q
_{i}(T) = g_{i} - q
_{i}(T) = 0

Now the question is if we try to interpolate the trajectory of q_{i}, we should use a cubic polynomial of *t*q_{i}(t) = a_{0} + a_{1}t + a_{2}t^{2} + a_{3}t^{3}

because we can determine the 4 coefficients a_{i}, i=0,1,2,3 from the above 4 constraints. The following values for the coefficients can be easily derived:

- a
_{0} = s_{i} - a
_{1} = 0 - a
_{2} = (3/T^{2}) * (g_{i} - s_{i}) - a
_{3} = (2/T^{3}) * (g_{i} - s_{i})

I don't understand how the a_{0},a_{1},a_{2},a_{3} are being derived. The things I have considered are:

For a_{0}: Are we setting t=0? So then we have: s_{i} = a_{0} + a_{1}(0) + a_{2}(0) + a_{3}(0).

For a_{1}: We have s_{i} = s_{i} + a_{1}t + a_{2}(0) + a_{3}(0) ? I don't think that is right, but it's the only way I see to set it up so that a_{1}=0.

For a_{2}: I'm pretty lost here. I see that q_{i} - s_{i} is the displacement...but how are we getting 3/T^{2}??

For a_{3}: Same problem as with a_{2}

Can anyone help me out here? I tried to summarize the handout text some, but I can try to copy more of it if the information is insufficient. Any help is appreciated.