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Accelerating 1g for one hour.


That means it is accelerating 9.86m/s/s.


If we assume that it's relative motion is 0 at the start of this hour then that's as simple as multiplying the number of seconds out:

3600 seconds * 9.86m/s/s = 35,496 m/s is the final velocity. To accelerate, negatively (as they say in physics...but we commonly call it "deceleration", it's as simple as it being -9.86m/s/s for one hour.


So there's the first portion.



*reads the rest of the post*


Edit -


Yes, there is a formula for figuring out the distance traveled through the acceleration vectors (both the positive and negative vectors WILL add to the total distance traveled, but, thankfully, once you figure out one, you just multiply that by 2 to figure out the other.)


So, first, figure out total distance traveled for just the acceleration vector.


Double edit -

So, if I remember the formula correclty (didn't have much googling), it's vt + (a*t^2)/2 = d

d = distance traveled

v = initial velocity.

Since the initial velocity is 0, we can throw that out.


Doing the math, that's 63,892,800 meters.

So that's how long it traveled during it's acceleration vector.

Double it because it wil be exactly the same through the negative acceleration vector.

So that's 63892800*2 = 127,785,600 m

So, now we are 2/3 the way done. We need to find out how far it traveled during that one week period.

triple edit -

K. did the math. It works out.


I converted one week to seconds, to keep like terms.

7 days = 168 hours

1 hour = 3600 seconds

3600*168 = 604800 seconds

It is traveling at 35,496 m/s for 604800 seconds.

Mutiply that out:

That's 35,496 m/s * 604800 seconds = 21,467,980,800 meters

Now, add your acceleration vectors in:

21,467,980,800 m + 127,785,600 m = 21,595,766,400 meters traveled.


There's your answer.


Did show enough work for you or do you want me to be more thorough?

Originally posted by dadudemon

So, first, figure out total distance traveled for just the acceleration vector.

This is my question. How do I figure out the total distance traveled for just the acceleration vector?

Originally posted by Lucius
This is my question. How do I figure out the total distance traveled for just the acceleration vector?

I edited 3 times so you could see my work, each step of the way. Check out my post, again.

Edit - But, I couldn't find the formula, online, so I was going by memory. I may be wrong. Double check using google.

Double edit -

YEA! I finally found it: I was right, it's the square of time.

Here it is:

http://www.physicsclassroom.com/class/1dkin/u1l1e.cfm

Originally posted by dadudemon
I edited 3 times so you could see my work, each step of the way. Check out my post, again.

Edit - But, I couldn't find the formula, online, so I was going by memory. I may be wrong. Double check using google.

Double edit -

YEA! I finally found it: I was right, it's the square of time.

Here it is:

http://www.physicsclassroom.com/class/1dkin/u1l1e.cfm

Sorry, I didn't see the edited sections until after I posted.

Thanks, this is what I was looking for.

9.81m/s/s is a closer approximation of Earth's gravity



As for finding "how far" it traveled in physics it's called displacement.

I'm sure you just care about the equation (it's at the bottom) but the derivation is sort of enlightening. I always like to see where knowledge comes from.

We have one unknown for that part of the problem: displacement.
We have two knowns: acceleration and time.
We can derive one more piece of information easily: velocity (as dadudemon showed you)

We can't work with acceleration yet (that's the whole problem!) so we only have two things to work with. Because of that we can put them problem on a graph and turn it into geometry.

Time will be the x axis and velocity will by the y axis.

Acceleration is constant so you get a nice diagonal line if you graph them against each other. It looks like a triangle but more generally it is a trapezoid (because you could have started acceleration when already moving).

The area of a trapezoid is .5(b_1 + b_2)h. But our particular trapezoid in on that graph so we can put the information of the graph into the formula and write it as .5(v_i + v_f)t. v_i is the "initial velocity", v_f is the "final velocity" and t is the time it took to go from one to the other. And rather than area the we're now figuring out displacement! (we're most of the way there)

Remembering that "final velocity is acceleration times time plus initial velocity" (which dadude showed) we can rewrite the equation to say:

.5(v_i + at + v_i)t

You could use this and get the right answer but it's messy so it's rearranged as:

d = v_i * t + .5at^2
-or-
displacement equals initial velocity times one half the time spent traveling plus acceleration times the square of the time of spent traveling.

Explanation from:
ttp://zonalandeducation.com/mstm/physics/mechanics/kinematics/EquationsForAcceleratedMotion/Origins/Displacement/Origin.htm

Originally posted by Symmetric Chaos
9.81m/s/s is a closer approximation of Earth's gravity



As for finding "how far" it traveled in physics it's called displacement.

I'm sure you just care about the equation (it's at the bottom) but the derivation is sort of enlightening. I always like to see where knowledge comes from.

We have one unknown for that part of the problem: displacement.
We have two knowns: acceleration and time.
We can derive one more piece of information easily: velocity (as dadudemon showed you)

We can't work with acceleration yet (that's the whole problem!) so we only have two things to work with. Because of that we can put them problem on a graph and turn it into geometry.

Time will be the x axis and velocity will by the y axis.

Acceleration is constant so you get a nice diagonal line if you graph them against each other. It looks like a triangle but more generally it is a trapezoid (because you could have started acceleration when already moving).

The area of a trapezoid is .5(b_1 + b_2)h. But our particular trapezoid in on that graph so we can put the information of the graph into the formula and write it as .5(v_i + v_f)t. v_i is the "initial velocity", v_f is the "final velocity" and t is the time it took to go from one to the other. And rather than area the we're now figuring out displacement! (we're most of the way there)

Remembering that "final velocity is acceleration times time plus initial velocity" (which dadude showed) we can rewrite the equation to say:

.5(v_i + at + v_i)t

You could use this and get the right answer but it's messy so it's rearranged as:

d = v_i * t + .5at^2
-or-
displacement equals initial velocity times one half the time spent traveling plus acceleration times the square of the time of spent traveling.

Explanation from:
ttp://zonalandeducation.com/mstm/physics/mechanics/kinematics/EquationsForAcceleratedMotion/Origins/Displacement/Origin.htm


WTF?

Where did I get 9.86 from?


Edit - lulz. I figured out why. Precision. 9.806 m/s/s I must have dropped a digit of precision through a shitty memory.

Originally posted by Symmetric Chaos
9.81m/s/s is a closer approximation of Earth's gravity



As for finding "how far" it traveled in physics it's called displacement.

I'm sure you just care about the equation (it's at the bottom) but the derivation is sort of enlightening. I always like to see where knowledge comes from.

We have one unknown for that part of the problem: displacement.
We have two knowns: acceleration and time.
We can derive one more piece of information easily: velocity (as dadudemon showed you)

We can't work with acceleration yet (that's the whole problem!) so we only have two things to work with. Because of that we can put them problem on a graph and turn it into geometry.

Time will be the x axis and velocity will by the y axis.

Acceleration is constant so you get a nice diagonal line if you graph them against each other. It looks like a triangle but more generally it is a trapezoid (because you could have started acceleration when already moving).

The area of a trapezoid is .5(b_1 + b_2)h. But our particular trapezoid in on that graph so we can put the information of the graph into the formula and write it as .5(v_i + v_f)t. v_i is the "initial velocity", v_f is the "final velocity" and t is the time it took to go from one to the other. And rather than area the we're now figuring out displacement! (we're most of the way there)

Remembering that "final velocity is acceleration times time plus initial velocity" (which dadude showed) we can rewrite the equation to say:

.5(v_i + at + v_i)t

You could use this and get the right answer but it's messy so it's rearranged as:

d = v_i * t + .5at^2
-or-
displacement equals initial velocity times one half the time spent traveling plus acceleration times the square of the time of spent traveling.

Explanation from:
ttp://zonalandeducation.com/mstm/physics/mechanics/kinematics/EquationsForAcceleratedMotion/Origins/Displacement/Origin.htm

My level of math comprehension is somewhere around what most colleges would call Math 99 (just short of intermediate algebra) so I'm in no way sure I actually understand this, but I think I actually get it. Thanks, I do like understanding the actual principles behind the formulas.

Originally posted by Lucius
My level of math comprehension is somewhere around what most colleges would call Math 99 (just short of intermediate algebra) so I'm in no way sure I actually understand this, but I think I actually get it. Thanks, I do like understanding the actual principles behind the formulas.

The version at the bottom is better. They have actual graphics and proper formatting for the equations.

Originally posted by dadudemon
Accelerating 1g for one hour.


That means it is accelerating 9.86m/s/s.


If we assume that it's relative motion is 0 at the start of this hour then that's as simple as multiplying the number of seconds out:

3600 seconds * 9.806 m/s/s = 35,301 m/s is the final velocity. To accelerate, negatively (as they say in physics...but we commonly call it "deceleration", it's as simple as it being -9.86m/s/s for one hour.


So there's the first portion.



*reads the rest of the post*


Edit -


Yes, there is a formula for figuring out the distance traveled through the acceleration vectors (both the positive and negative vectors WILL add to the total distance traveled, but, thankfully, once you figure out one, you just multiply that by 2 to figure out the other.)


So, first, figure out total distance traveled for just the acceleration vector.


Double edit -

So, if I remember the formula correctly (didn't have much googling), it's vt + (a*t^2)/2 = d

d = distance traveled

v = initial velocity.

Since the initial velocity is 0, we can throw that out.


Doing the math, that's 63,542,880 meters.

So that's how long it traveled during it's acceleration vector.

Double it because it wil be exactly the same through the negative acceleration vector.

So that's 63,542,880*2 = 127,085,760 m

So, now we are 2/3 the way done. We need to find out how far it traveled during that one week period.

triple edit -

K. did the math. It works out.


I converted one week to seconds, to keep like terms.

7 days = 168 hours

1 hour = 3600 seconds

3600*168 = 604800 seconds

It is traveling at 35,301 m/s for 604800 seconds.

Multiply that out:

That's 35,301 m/s * 604800 seconds = 21,350,044,800 meters

Now, add your acceleration vectors in:

21,350,044,800 m + 127,085,760 m = 21,477,130,560 meters traveled.


There's your answer.


Did show enough work for you or do you want me to be more thorough?

I edited the above post for the correct g.

g = 9.806

Yes, I eliminated some precision, unsoundly.

I hate discussing math on forums. Trying to type out algebraic notation is a complete nightmare.

Originally posted by King Kandy
I hate discussing math on forums. Trying to type out algebraic notation is a complete nightmare.

There are ways to do it (superscript and subscript) with key combinations but they do not work on forums.


You can still do things like

f(x)=ax+b



But, you're right, it's much better if you have "active" instead of tagged formatting. Some forums actually do have that for their posting platform and it's quite nice.

Some maths forums I have been to had formula creators.

Red leader, standing by!