# Thread: Converting Accelleration to Velocity

1. ## Converting Accelleration to Velocity

I need to convert accelleration to velocity.

With an accelleration of 1.32944 x 10^(-7) m/s^2, and an innitial velocity of 0 m/s, to find the velocity, I would need to find the square root of the accelleration, then multiply by the time at that velocity, correct?

2. Originally Posted by Thrawn
I need to convert accelleration to velocity.

With an accelleration of 1.32944 x 10^(-7) m/s^2, and an innitial velocity of 0 m/s, to find the velocity, I would need to find the square root of the accelleration, then multiply by the time at that velocity, correct?
The accelration formula is this:

a = (v_f - v_0)/t

You can rewrite it to look like this:

v_f = v_0 + at

v_f = (1.32944 x 10^(-7))t

All you need is JUST the time, you don't need to find the square root of anything.

To prove it can actually be written that way is this:

You have a position-time function x(t):

x(t) = x_0 + v_0t + 1/2at^2

We know that the derivative of x(t) = v(t) or a velocity-time function

The derivative of the constant x_0 = 0
The derivative of v_0t = v_0 because the derivative of t^n = nt^n-1, so if v_0t then 1 * v_0 = 1v_0 = v_0 and t^(1-1) = t^0 = 1: v_0 * 1 = v_0
The derivative of 1/2at^2 = at because the derivative of t^n = nt^n-1, so if 1/2at^2 then 2 * 1/2a = 1a = a and t^(2-1) = t^1 = t: a * t = at

v(t) = v_0 + at

Thus proving our formula is correct where v(t) = v_f

Another way is to take the given acceleration formula:

a = (v_f - v_0)/t

multiply by t on both sides:

at = v_f - v_0

at + v_0 = v_f

This ALSO proves the formula above.

There are no radicals involved in this.

3. What would I need to do if I don't know the time?

4. Originally Posted by Thrawn
What would I need to do if I don't know the time?
You could use this:

x = x_0 + v_0t + 1/2 at^2

total distance traveled = (initial position) + 1/2 (1.32944 x 10^(-7))t^2

Solve for t if you know the initial position and total distance traveled, one of these is most likely zero.