Question

In the position formula x(t)=At2 + Bt + C, what is the value of the acceleration in terms of A?

Answer 1

Am I solving for A? Deriving?

Answer 2

Is this calculus level?

Answer 3

If this is calculus, then the acceleration is the second derivative of the position with respect to time.

Answer 4

What he said, yes.
If not I believe you have to isolate the A.

Answer 5

I think so, the class is physics.

Answer 6

Hmm, I wouldn't know then; my Physics teacher said the only necessary prerequisite for Physics is Pre-Calculus

Answer 7

Well, you can kind of make sense here, as of:

\(\color{#000000}{ \displaystyle x(t)=At^2 + Bt + C\quad \Longleftrightarrow \quad x(t)=\frac{1}{2}at^2+(v_{0x})t +x_0 }\)

Answer 8

Compare the two, and it should clearly follow that:
\(\color{#000000}{ \displaystyle At^2 =\frac{1}{2}at^2 }\)

Answer 9

where lower case a is acceleration.

Answer 10

We've already been using concepts from calc. The current subject is acceleration due to gravity on an inclined plane. Solomon's idea might work

Answer 11

You get the same thing either based on using the formula for displacement as a comparison, or if you differentiate twice.

Answer 12

I just noticed that O.o