# Homework Help: Changing a Formula

1. Sep 29, 2004

### TheShapeOfTime

[SOLVED] Changing a Formula

How do I switch the following formula for time?

delta d = Vi * delta t + 1/2 * a * delta t^2

d = distance
Vi = initial velocity
a = acceleration
t = time

Any sort of steps would be greatly appreciated as I don't just want the answer, I always like to know how it's done :).

P.S. Sorry about the wording of the question and how the formula is written out, I need to get better at it.

2. Sep 29, 2004

### arildno

What are the roots of a quadratic equation..?

3. Sep 29, 2004

### TheShapeOfTime

Sorry, but I don't really get what you mean (noobness :\). All I can tell you is that I know you solve quadratic equations by factoring.

EDIT: I removed my nonsense ;)

Last edited by a moderator: Sep 29, 2004
4. Sep 29, 2004

### arildno

You know how to solve for x the following equation, right?:
ax^2+bx+c=0 (a,b,c constants)

5. Sep 29, 2004

### TheShapeOfTime

Looks like you replied while I was editing my post. My answer is yes.

6. Sep 29, 2004

### arildno

It seems you have severe problems with simple algebra (your edit is sheer nonsense).
Now, in order to proceed, try to rewrite your original equation into a form:
$$K_{1}(\delta{t})^{2}+K_{2}(\delta{t})+K_{3}=0$$
where $$K_{1},K_{2},K_{3}$$ are constants (identify them!)
($$\delta{t}$$ is "delta t")

7. Sep 29, 2004

### TheShapeOfTime

$$K_{1} = 1/2a, K_{2} = Vf, K_{3} = d$$

Also, wouldn't it be $$-K_{3}$$, or am I wrong?

8. Sep 30, 2004

### arildno

What's wrong with setting:
$$K_{3}=-\delta{d}$$
Now, you should be able to determine which values $$\delta{t}$$
Choose the root which is makes sense physically.

9. Sep 30, 2004

### TheShapeOfTime

I understand now, thanks for helping me out!

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