quadratic formula

The quadratic formula tells you that a t^2 + b t + c = 0 if, and only if, t = [ -b +- sqrt(b^2 - 4 a c) ] / ( 2 a ).

You would apply the formula to solve the equation

`ds = v0 `dt + 1/2 a `dt^2

for `dt, as follows:

Rearrange the equation to the form

1/2 a `dt^2 + v0 `dt - `ds = 0.

substitute your values of a, v0 and `ds.

Whatever quantity you end up with in front of `dt^2 takes the role of a in the quadratic formula, not to be confused with the acceleration a (even though in this case the two are related).

Whatever quantity ends up in front of `dt takes the role of b.

Whatever - `ds happens to be, that's c.

Plug those numbers into the formula. Depending on the value of the discriminant, which is the b^2 - 4 a c quantity within the square root, you will get no solution (if b^2 - 4 a c is negative you can't take the square root), 1 solution (if b^2 - 4 a c = 0 then the + and the - of the +- give you the same thing) or 2 solutions (if b^2 - 4 a c is positive then its square root is real and nonzero, so you will get a solution for the + and a solution for the - of the +-).

If you get no solutions, then that combination of `ds, v0 and a is not possible.

If you get 1 or 2 solutions, then you have to decide which, if any, actually fit the problem.

As I said in class, you generally avoid solutions that require the quadratic formula unless you are left with no choice. In Phy 231 and 241, this happens frequently; in 201 is happens but only on a few problems in the entire course; in 121 it's not something you have to worry about.

In any situation where you have a, v0 and `ds you can instead use the fourth equation vf^2 = v0^2 + 2 a `ds to solve for vf, then you can just use straightforward reasoning to find `dt.