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## Homework Statement

This isn't really a homework question that I have but more of a question i am trying to find out myself. When objects are near the surface of the earth, objects accelerate at an average of 9.8 meters per second^2. However when we get farther away from the earth, say have the distance it takes to get to the moon, the acceleration is different. Not only that but as the object gets closer to the earth the acceleration increases and can not be treated as a constant anymore. I've been thinking in order to resolve this I must try to make an equation with change in acceleration, a jerk. I want to see what you guys think there are any flaws in my process. I also want to figure out the amount of time from my final equation as the title implies.

## Homework Equations

[tex] A= \frac{\ a+ao}{2}[/tex]

[tex] j= \frac{\ a-ao}{t}[/tex]

[tex] d= vot+.5At^2[/tex]

Where A is the average acceleration, a is the final acceleration, ao is the initial acceleration, j is the average jerk, t is time, d is the amount of displacement and vo is the initial velocity.

## The Attempt at a Solution

I start with this equation-

[tex] d= vot+.5At^2[/tex]

I notice that it has the variable A which needs to be taken out. So I replace A with this equation-

[tex] A= \frac{\ a+ao}{2}[/tex]

and it becomes

[tex] d= vot+.5(\frac{\ a+ao}{2})t^2[/tex]

Next i need to decide to get rid of a or ao so i can somehow put j in there. It doesn't really matter but i decide to replace a with this equation

[tex] j= \frac{\ a-ao}{t}[/tex]

[tex] jt= a-ao[/tex]

[tex] jt+ao=a[/tex]

So once again the new equation is this-

[tex] d= vo+.5(\frac{\ jt+ao+ao}{2})t^2[/tex]

Now i go through the equation and simplify

[tex] d= vot+.5(\frac{\ jt+2ao}{2})t^2[/tex]

[tex] d= vot+.5(\frac{\ jt^3+2aot^2}{2})[/tex]

[tex] d= vot+(\frac{\ jt^3+2aot^2}{4})[/tex]

[tex] d= \frac{4vot}{4}+\frac{\ jt^3+2aot^2}{4}[/tex]

[tex] d= \frac{\ jt^3+2aot^2+4vot}{4}[/tex]

this is what i get for a final equation

[tex] d= \frac{\ jt^3+2aot^2+4vot}{4}[/tex]

This equation makes sense because when j is 0 you get the original equation of

[tex] d= vot+.5At^2[/tex]

and ao becomes A because there is no change in acceleration.

However I don't know if my final equation is correct for sure and I want some feedback to see if it is right. If it is right then we can move on to the question of finding out what t is when all other variables are known. The problem is I don't know how to solve for t. My mind says to make one side of the equation 0 so

[tex] 0= \frac{\ jt^3+2aot^2+4vot-4d}{4}[/tex]

And i guess i can get rid of the 4 in the denominator by multiplying both sides of the equation by 4, giving me this

[tex] 0= jt^3+2aot^2+4vot-4d[/tex]

But then what? I can't use the quadratic formula. I could factor like this

[tex] 0= t(jt^2+2aot+4vo)-4d[/tex]

but that doesn't really get me to far. Can someone help, please?

Thanks for helping me out.