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## iwanttogotostanford one year ago I

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1. iwanttogotostanford

@zepdrix @triciaal

2. iwanttogotostanford

I need to know the acceleration due to gravity and the average G value for each of them given what I have

3. iwanttogotostanford

E Earth Moon Mars Distance (m) 2.25 3.08 4.04 4.57 4.68 3.55 Time (s) 0.680 0.793 2.20 2.26 1.58 1.38 Acceleration due to Gravity (m/s^2) 9.732 9.795 1.669 1.789 3.749 3.728 Average G value for the location (m/s^2)

4. iwanttogotostanford

I NEED TO KNOW THE AVERAGE G VALUE FOR THE LOCATION

5. iwanttogotostanford

I need the equation or formula please I am confused

6. anonymous

a picture of the chart/table will work better methinks but the acceleration could be obtained from the slope I'd think

7. anonymous

and average... is more or less rate of change, which brings us back to slope again

8. iwanttogotostanford

Earth Moon Mars Distance (m) 2.25 3.08 4.04 4.57 4.68 3.55 Time (s) 0.680 0.793 2.20 2.26 1.58 1.38 Acceleration due to Gravity (m/s^2) 9.732 9.795 1.669 1.789 3.749 3.728 Average G value for the location (m/s^2)

9. anonymous

then again, a picture of the table woiuld help :)

10. iwanttogotostanford

This is the chart but it looks a little messed up but it is in line

11. anonymous

well, surely doesn't look like that on your screen... so just screenshot it methink

12. iwanttogotostanford

I'm on a macbook

13. iwanttogotostanford

but I will try

14. iwanttogotostanford

15. iwanttogotostanford

Here it is

16. iwanttogotostanford

it is a download did you get it

17. anonymous

http://prntscr.com/8q0dt3 There's a screenshot if you can't download the document for some strange reason. For future reference, press command + shift + 4 to screenshot something on a Mac.

18. anonymous

yes.... well you have two values per each location so the average "G"ravity will be the sum of both, divided by 2 say for earth will be $$\bf \cfrac{9.732+9.795}{2}$$

19. iwanttogotostanford

So would they be in order= 14.6295 2.5635 5.613

20. iwanttogotostanford

@jdoe0001

21. iwanttogotostanford

@jdoe0001 Please check

22. iwanttogotostanford

@triciaal Did i do the table right?? :-)

23. triciaal

no idea haven't look yet

24. iwanttogotostanford

@triciaal could you check please

25. triciaal

not sure of final for each location same process velocity = distance /time acceleration = velocity/time for Earth (3.08 - 2.25)/ (0.793-0.680) = velocity velocity/time(1) = 9.732 velocity/time(2) = 9.795 average G = velocity/(t2 - t1)

26. triciaal

|dw:1444528438601:dw|

27. triciaal

@mathmate can you help with this?

28. triciaal

@jim_thompson5910 ?

29. mathmate

@iwanttogotostanford You would need the kinematics equation for an object in free fall, $$\Delta x= (1/2)gt^2$$ where g=acceleration due gravity, t=time it takes, (initial velocity = 0 for free fall) Solve for g to get $$g=2\Delta x /t^2$$ That's how @jdoe001 got 9.732 and 9.795 for the first pair of data. Continue with the two other pairs for moon and mars, and you'll have a nice table.

30. mathmate

Average value would be, as @jdoe0001 stated, the sum divided by 2, for each pair.

31. iwanttogotostanford

@mathmate I know but what about for the last row of the table- i am confused

32. iwanttogotostanford

@triciaal

33. iwanttogotostanford

@jim_thompson5910 how do you figure out the last row i know the third row but did i do the last row correctly?

34. mathmate

Location Earth Moon Mars Acceleration due to Gravity (m/s^2) $$\color{red}{(}$$9.732 9.795$$\color{red}{)}$$ $$\color{red}{(}$$1.669 1.789$$\color{red}{)}$$ $$\color{red}{(}$$3.749 3.728$$\color{red}{) }$$ Average acceleration 9.764 1.729 3.739 For earth, the average is (9.732+9.795)/2=9.764 For moon, the average is (1.669+1.789)/2=1.729 For mars, the average is (3.749+3.728)/2=3.739

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