Knuckle dragger with question about acceleration

[rojer ramjet]

Howdy guys!

First I should probably introduce myself and give a little educational history, not necessarily so y'all can take pity on me and do my work for me, but so that you can understand the purpose of my question - I'd like to try and figure it out myself; I feel a great sense of accomplishment in figuring things out.

Anyway, I'm a high-school dropout who spent 22 years in the Army; I've accrued more than 200 credit hours in community college, just taking classes that interest me; I'm also a licensed Master Electrician, AWS Certified Critical Members and Structural Steel Welder, an amateur ballistician (I used to own a firearms and ammunition manufacturing company, for which I did all design and testing); currently I do commercial salvage, search and recovery, and public safety diving; I have contracts with the DOD, Coast Guard and several Sheriff departments in California, Oregon and Washington State. I'm distantly familiar with gas physics.

I'm not a dummy, I'm just uneducated.

I'm also a HUGE science fiction fan; my favorite is fiction based upon Newtonian physics; speed of light is absolute, anti-matter is rare and not used as a power source, etc.

My question: In many of the books I've read, such as from Clarke, Reynolds, Ctein, Corey, etc, acceleration has been described in percentages of "G," instead of "miles per hour," things under constant acceleration could only easily be described like that, right?

I'd like to understand the "relative velocity," but I don't want to make a graph to determine what velocity a body would be traveling after a certain period of known acceleration, which is the only way I know how to calculate such large numbers - is there an equation that would allow me to calculate relative velocity based upon acceleration according to the speed of gravity on Earth? Is that even the right way to frame the question? lol

And guidance in finding the answer for myself would be appreciated much more than the answer - I can do algebra well, very well; I tutor in math A and D at Sierra College, though I'm afraid I'm a victim of "trained monkey syndrome;" I know how to get the answer, I'm just not sure why I use the steps that I've been trained to use to find the answer. I have minimal skills in trigonometry and geometry, and none in calculus or differential calculus.

Warm regards!

 

[anorlunda]

Einstein showed us that the force of acceleration feels just like the force of gravity. We can't tell the difference.

If you weigh 150 pounds on Earth, and if you stand on a scale in a spaceship accelerating at 1G, the scale reads 150 pounds.

So the G is not a measure of velocity, it is a measure of acceleration.

 

[A.T.]

is there an equation that would allow me to calculate relative velocity based upon acceleration

https://en.wikipedia.org/wiki/Equat...translational_acceleration_in_a_straight_line

 

[rojer ramjet]

Einstein showed us that the force of acceleration feels just like the force of gravity. We can't tell the difference.

If you weigh 150 pounds on Earth, and if you stand on a scale in a spaceship accelerating at 1G, the scale reads 150 pounds.

So the G is not a measure of velocity, it is a measure of acceleration.

Correct; my question was "how do I calculate relative velocity based upon known acceleration." I think A.T. pointed me in the right direction. Thanks!

 

[berkeman]

how do I calculate relative velocity based upon known acceleration

The acceleration of each object can help you calculate the velocity versus time for that object. Then when you have the two velocities as a function of time, you just subtract them to get the relative velocity.

In general each velocity will be a vector, and you do the subtraction with vectors in 3-dimensional space. If you can simplify the problem to linear velocities along the same line, then that simplifies into just subtracting their linear speeds.

Hope that helps.

 

[fresh_42]

My question: In many of the books I've read, such as from Clarke, Reynolds, Ctein, Corey, etc, acceleration has been described in percentages of "G," instead of "miles per hour," things under constant acceleration could only easily be described like that, right?

No. Miles per hour measures a speed, an acceleration is a change of speed in a certain time period, so we would measure it as miles per hour per hour. $G$ is roughly $10$ meters per second per second, or $10$ meters per square second: $10 \frac{\frac{m}{s}}{s}=10 \frac{m}{s^2}$. It means if we jump from an airplane, where we have zero velocity towards the ground, we will have a speed of ten meters per second after one second of free fall, then 20 meters per second after two seconds of free fall etc. - of course disregarding air resistance. In real life there is a final velocity in air upon which there is no more acceleration. That's why parachuting works.

So although $G$ represents a constant acceleration of ten meters per second per second, it is still an acceleration, that's a change of velocity, namely by ten meters per second more every second. So an acceleration of say $5\, G$ means, that if there wasn't a boundary as in a jet cockpit, the accelerated person would have a speed of $50\,\frac{m}{s}= 112\,mph$ in just one second! And $224\,mph$ after two seconds. So the $G-$scale is a bit more convenient than to handle increasing velocities in a short time, or the more or less abstract notation of $50$ meters per square second. $5\,G$ is simply five times the weight we usually have due to acceleration, resp. push at a boundary as a jet cockpit or a roller coaster car.

 

[rojer ramjet]

This is kinda what I was looking for. Now I just gots to figure out the equation; is PEMDAS the correct way to look at this problem, or "the other way?"

 

[rojer ramjet]

Hope that helps.

73's!

 

[fresh_42]

View attachment 222493

This is kinda what I was looking for. Now I just gots to figure out the equation; is PEMDAS the correct way to look at this problem, or "the other way?"

This looks like my example with the airplane: The plane travels at a constant speed of $v_0$, $a$ would be the acceleration after leaving the airplane, i.e. $a=G$ on earth, and $r(t)$ is the location of the jumper at time $t$.

 

[rojer ramjet]

No. Miles per hour measures a speed, an acceleration is a change of speed in a certain time period, so we would measure it as miles per hour per hour. $G$ is roughly $10$ meters per second per second, or $10$ meters per square second: $10 \frac{\frac{m}{s}}{s}=10 \frac{m}{s^2}$. It means if we jump from an airplane, where we have zero velocity towards the ground, we will have a speed of ten meters per second after one second of free fall, then 20 meters per second after two seconds of free fall etc. - of course disregarding air resistance. In real life there is a final velocity in air upon which there is no more acceleration. That's why parachuting works.

So although $G$ represents a constant acceleration of ten meters per second per second, it is still an acceleration, that's a change of velocity, namely by ten meters per second more every second. So an acceleration of say $5\, G$ means, that if there wasn't a boundary as in a jet cockpit, the accelerated person would have a speed of $50\,\frac{m}{s}= 112\,mph$ in just one second! And $224\,mph$ after two seconds. So the $G-$scale is a bit more convenient than to handle increasing velocities in a short time, or the more or less abstract notation of $50$ meters per square second. $5\,G$ is simply five times the weight we usually have due to acceleration, resp. push at a boundary as a jet cockpit or a roller coaster car.

Thanks for the explanation. :)

 

[Dr.D]

It helps to start with a few definitions.

Position is where something is located. It is measured as a distance from a reference mark, in perhaps feet, inches, meters, or millimeters.

Velocity is how fast position is changing. It is measured in speed units, such as ft/sec, meters/sec, mi/hr, km/hr, etc.

Acceleration is how fast velocity is changing. It is measured in units that talk about change in speed per unit of time, such as inches/sec^2, ft/s^2, km/s^2, etc.

g stands for a physical constant called the acceleration of gravity. The value of g depends upon what units system you want to use. Typical values are
g = 981 cm/s^2 in CGS units
g = 9.807 m/s^2 in SI units
g = 32.174 ft/s^2 in Foot-Pound-Second (FPS) units
g = 386.088 in/s^2 in Inch-Pound-Second (IPS) units

 

[russ_watters]

Welcome to PF and thanks for your service!

My question: In many of the books I've read, such as from Clarke, Reynolds, Ctein, Corey, etc, acceleration has been described in percentages of "G," instead of "miles per hour," things under constant acceleration could only easily be described like that, right?

In addition to the answers given, I'd also say that it is more concise to say "2.5g" instead of "55 miles per hour per second". The units, when spoken-out, are cumbersome.

 

[waves and change]

Howdy guys!

First I should probably introduce myself and give a little educational history, not necessarily so y'all can take pity on me and do my work for me, but so that you can understand the purpose of my question - I'd like to try and figure it out myself; I feel a great sense of accomplishment in figuring things out.

Anyway, I'm a high-school dropout who spent 22 years in the Army; I've accrued more than 200 credit hours in community college, just taking classes that interest me; I'm also a licensed Master Electrician, AWS Certified Critical Members and Structural Steel Welder, an amateur ballistician (I used to own a firearms and ammunition manufacturing company, for which I did all design and testing); currently I do commercial salvage, search and recovery, and public safety diving; I have contracts with the DOD, Coast Guard and several Sheriff departments in California, Oregon and Washington State. I'm distantly familiar with gas physics.

I'm not a dummy, I'm just uneducated.

I'm also a HUGE science fiction fan; my favorite is fiction based upon Newtonian physics; speed of light is absolute, anti-matter is rare and not used as a power source, etc.

My question: In many of the books I've read, such as from Clarke, Reynolds, Ctein, Corey, etc, acceleration has been described in percentages of "G," instead of "miles per hour," things under constant acceleration could only easily be described like that, right?

I'd like to understand the "relative velocity," but I don't want to make a graph to determine what velocity a body would be traveling after a certain period of known acceleration, which is the only way I know how to calculate such large numbers - is there an equation that would allow me to calculate relative velocity based upon acceleration according to the speed of gravity on Earth? Is that even the right way to frame the question? lol

And guidance in finding the answer for myself would be appreciated much more than the answer - I can do algebra well, very well; I tutor in math A and D at Sierra College, though I'm afraid I'm a victim of "trained monkey syndrome;" I know how to get the answer, I'm just not sure why I use the steps that I've been trained to use to find the answer. I have minimal skills in trigonometry and geometry, and none in calculus or differential calculus.

Warm regards!

Keep in mind you mentioned known acceleration. The problem is less trivial if it’s a time varying acceleration ! Constant acceleration was explained by others in the thread...

 

[rojer ramjet]

Wowsers! So much information! Thanks a lot!

 

[rojer ramjet]

Welcome to PF and thanks for your service!

In addition to the answers given, I'd also say that it is more concise to say "2.5g" instead of "55 miles per hour per second". The units, when spoken-out, are cumbersome.

Thank you for your support; it was an honor and a privilege to serve my nation in good times and bad.