# Acceleration Distance (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

#### bhughes01

Can someone help? I am trying to find a formula to solve the following problem:

What distance would a body have to travel to attain a speed of 25,000 an hour with the acceleration limited to producing a given number of g-forces on the body, say 4g.

So the acceleration would produce a constant g-force of 4 on the body.

The desired speed is 25,000 MPH.

How far would the body have to travel to attain it?

The twist in the calculation is using g-force as an acceleration measurement.

I am trying to solve a problem in my design of a theoretical space transportation system. I need that acceleration length as a vital piece of the puzzle.

You can find my design at http://www.bhughes.com/ "Space Travel"

Thanks,

Brian L Hughes

#### Hootenanny

Staff Emeritus
Gold Member
Hi there Brian and welcome to PF,

What you need is a kinematic formula for constant acceleration. However, in addition to the information provided you will also require a time (for the acceleration) or an initial velocity, which can be zero or otherwise. What formula you would use is dependant upon the other parameters. For example, if you also know the intial velocity you could use the equation; $v^{2} = u^{2} + 2as$, where v is your final velocity, u your intial velocity (this can be zero if your object is starting from rest), a is your acceleration and s is the distance travelled. Alternatively, if you know the time taken from the acceleration you could use $s = ut + \frac{1}{2}at^{2}$, where t is time.

A word of caution, be careful with your units here. Hope this helps.

#### bhughes01

what are my "units" that I need to be careful of? Is there some kind of secret code used here? At least I'm not typing in all caps. :O)

I know my target speed. I know the max g-force that I am willing to take. The initial velocity is zero.

for example:

speed=25,000 mph
gforce=4g
what distance=feet, miles, or meters?

so first I need a formula for g-force so I can figure out the ceiling for my acceleration

once I know my maximum acceleration then I just calcuate how far I have to travel at that acceleration to reach my goal of 25,000 per hour.

I've tried to find some formula on the www before and came up empty. I am not ashamed to declare my lack of knowledge of physics. I do know enough to at least realize I need formula.

#### Hootenanny

Staff Emeritus
Gold Member
bhughes01 said:
what are my "units" that I need to be careful of? Is there some kind of secret code used here? At least I'm not typing in all caps. :O)
The units I was refering to is the fact that your speeds are in miles per hour and acceleration due to gravity is usually quoted in meters per second, all your values must be in the same units.
bhughes01 said:
I know my target speed. I know the max g-force that I am willing to take. The initial velocity is zero.

for example:

speed=25,000 mph
gforce=4g
what distance=feet, miles, or meters?

so first I need a formula for g-force so I can figure out the ceiling for my acceleration

once I know my maximum acceleration then I just calculate how far I have to travel at that acceleration to reach my goal of 25,000 per hour.
4g is simply four times the acceleration due to gravity; 4g = 4 x 9.81 = 39.24 m.s-2, which in itself is an acceleration. g - force confusingly is actually a unit of acceleration not force. So in this case you need the first formula $v^2 = u^2 + 2as$, as your initial velocity is zero the u term drops out and we can rearrange the equation to become;

$$S = \frac{V^2}{2a}$$

bhughes01 said:
I've tried to find some formula on the www before and came up empty. I am not ashamed to declare my lack of knowledge of physics. I do know enough to at least realize I need formula.
There is no shame in lack of knowledge, only the stupid or the ignorant believe they know everything

#### denni89627

after a quick calculation i came up with 285.10 sec. I used formula
average acceleration = change in velocity/change in time Curious if it's right.

[EDIT] sorry i though you were looking for time. oops.

#### denni89627

OK i came up with 1593150m or approx 989 miles. not sure if im right.

#### bhughes01

Thanks, I've updated the document on my web site and credited the physics forum.

#### Hootenanny

Staff Emeritus
Gold Member
bhughes01 said:
Thanks, I've updated the document on my web site and credited the physics forum.
No problem, it was my pleasure

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving