# Solving for T in an acceleration equation

1. Mar 3, 2013

http://imgur.com/HeqIYqR

How do I solve for Δt when there's a Δt^2

2. Mar 3, 2013

### lep11

3. Mar 3, 2013

So what would b, a and c be?

4. Mar 3, 2013

### idllotsaroms

when you set the equation equal to zero a is the t^2 value, b is value with t variable, and c is just the number without the variable with it.

"ax^2 + bx + c"

5. Mar 3, 2013

Last edited: Mar 3, 2013
6. Mar 3, 2013

### idllotsaroms

I thought you were trying to solve for the change in time? So you would set the equation equal to zero d = d∅ + v∅t + 1/2 vt^2
and you just put d - d∅ = Δd
Δd = v∅t + 1/2 vt^2

where v∅=initial velocity

so the v∅t would be the b value
1/2 * v would be the a value
you would subtract Δd from both sides to get the equation equal to 0 and Δd would be your c value

it would look like
0 = 1/2vt^ + v∅t - Δd
0 = at^2 + bt + c

so you'll get two answers one at the max height of trajectory when velocity = 0 and one at the end of the trajectory where velocity = 0 again