Calculate Distance w/ 4g Acc. to Reach 25,000 MPH

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Discussion Overview

The discussion revolves around calculating the distance required for a body to reach a speed of 25,000 miles per hour under a constant acceleration that produces 4g of force. The participants explore the necessary formulas and considerations for this calculation, focusing on the implications of using g-force as a measurement of acceleration.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant seeks a formula to determine the distance needed to reach a target speed of 25,000 MPH with a constant acceleration of 4g.
  • Another participant suggests using kinematic equations, noting that additional parameters such as time or initial velocity are needed for the calculations.
  • There is a discussion about the importance of consistent units, particularly converting speeds from miles per hour to meters per second for accurate calculations.
  • A participant calculates the acceleration corresponding to 4g and provides a formula for distance based on the target velocity and acceleration.
  • One participant shares a calculation for time taken to reach the target speed but later realizes it was not the requested information.
  • Another participant provides a distance estimate based on their calculations but expresses uncertainty about its accuracy.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the calculations or the final distance required. Multiple views and calculations are presented, with some participants expressing uncertainty about their results.

Contextual Notes

Participants highlight the need for consistent units in calculations, particularly when converting between different measurement systems. There are unresolved aspects regarding the accuracy of the calculations and the assumptions made in the process.

Who May Find This Useful

This discussion may be useful for individuals interested in theoretical physics, space travel design, or those looking to understand the application of kinematic equations in practical scenarios.

bhughes01
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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
 
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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 dependent 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.
 
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.
 
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 referring 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}

Where V is your target velocity, a is your acceleration (in your example 39.2439.24 m.s-2) and S is your distance travelled.
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:smile:
 
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.
 
OK i came up with 1593150m or approx 989 miles. not sure if I am right.
 
Thanks, I've updated the document on my website and credited the physics forum.
 
bhughes01 said:
Thanks, I've updated the document on my website and credited the physics forum.
No problem, it was my pleasure :smile:
 

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