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

In summary, the conversation revolved around finding a formula to calculate the distance a body would have to travel in order to attain a speed of 25,000 miles per hour with a limited acceleration of 4 g-forces. The formula used was s = v^2/2a, where s is the distance, v is the target velocity, and a is the acceleration. The conversation also clarified the units that should be used and provided a calculation for the distance, which was approximately 989 miles.
  • #1
bhughes01
3
0
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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  • #2
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; [itex]v^{2} = u^{2} + 2as[/itex], 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 [itex]s = ut + \frac{1}{2}at^{2}[/itex], where t is time.

A word of caution, be careful with your units here. Hope this helps.
 
  • #3
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.
 
  • #4
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 [itex]v^2 = u^2 + 2as[/itex], as your initial velocity is zero the u term drops out and we can rearrange the equation to become;

[tex]S = \frac{V^2}{2a}[/tex]

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:
 
  • #5
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.
 
  • #6
OK i came up with 1593150m or approx 989 miles. not sure if I am right.
 
  • #7
Thanks, I've updated the document on my website and credited the physics forum.
 
  • #8
bhughes01 said:
Thanks, I've updated the document on my website and credited the physics forum.
No problem, it was my pleasure :smile:
 

1. How do you calculate distance with 4g acceleration to reach 25,000 MPH?

The formula for calculating distance with 4g acceleration is d = 0.5 * a * t^2, where d is distance, a is acceleration in meters per second squared, and t is time in seconds. To reach 25,000 MPH, you would need to convert the speed to meters per second (11,176.2 m/s) and then plug it into the formula along with the acceleration of 4g (39.2 m/s^2).

2. What is the acceleration needed to reach 25,000 MPH in a specific time?

The acceleration needed to reach 25,000 MPH in a specific time can be calculated by dividing the desired speed by the time it takes to reach that speed. For example, if you wanted to reach 25,000 MPH in 10 seconds, the acceleration needed would be 2,500 MPH/second or 1,117.62 m/s^2.

3. Can the distance be calculated without knowing the acceleration?

No, the distance cannot be accurately calculated without knowing the acceleration. The formula for distance incorporates acceleration, so without that information, the calculation would be incomplete.

4. How does 4g acceleration compare to other accelerations?

4g acceleration is equivalent to 39.2 meters per second squared. This is roughly four times the acceleration due to gravity on Earth (9.8 m/s^2) and is considered a high level of acceleration. It is commonly experienced in roller coasters and high-performance cars.

5. Is 4g acceleration safe for humans?

4g acceleration is generally considered safe for humans to experience for short periods of time. However, prolonged exposure to this level of acceleration can cause discomfort and potential health issues, such as blackouts or whiplash. Safety precautions should always be taken when experiencing high levels of acceleration.

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