# Constant acceleration and time

1. Sep 11, 2008

### Photon713

Is there a formula that will calculate time when I know the following...?

Initial velocity = 0 Final velocity = 28.14 feet per second and distance traveled = 3 feet.

I would like to know the formula and the time it would take to reach a velocity of
28.14 feet per second if I only have 3 feet to reach that velocity. I would like to
alter the velocity and re-calculate the time.

Thanks for any help...I have no physics or calculus background.

2. Sep 11, 2008

### tiny-tim

Welcome to PF!

Hi Photon713! Welcome to PF!

With constant acceleration, the average speed is half the maximum speed:

vaverage = vfinal/2.

Then time = distance/vaverage.

3. Sep 11, 2008

### HallsofIvy

Here's a trick that will help. If an object is moving with a constant acceleration, it "average velocity"- total distance moved, divided by time moved- is equal to the arithmetic average of the beginning and final velocities. In this case, Your average velocity 3/T, where T is the time required and since your final velocity is 28.14 and initial velocity is 0, the average velocity is (28.14+ 0)/2= 14.07 ft/sec. Now you have
3/T= 14.07 ft/sec so T= 3/14.07 seconds.

More generally, to go from velocity 0 to V in distance D, you have an average velocity of D/T= V/2 so T= 2D/V. Of course, that must be done at acceleration V/T.

4. Sep 11, 2008

### Photon713

Gentlemen...thank you for your responses. It's exactly what I wanted. Appreciate the welcome and the formula. Regards