Astronaut's Maximum Distance from Shuttle Due to Line Breakage

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SUMMARY

The discussion centers on calculating the maximum distance an astronaut can drift from a shuttle after her tether line breaks. The astronaut, weighing 63.1 kg, throws a 12.9 kg oxygen tank at a speed of 10.4 m/s to propel herself back. The correct final velocity (Vf) for the astronaut is calculated to be 2.126 m/s. Using the distance formula for constant velocity, the maximum distance she can be from the shuttle while still returning within 52.2 seconds is determined to be 111.912 m, not the incorrect 55.49 m initially calculated.

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Homework Statement


A 63.1 kg astronaut is on a space walk away from the shuttle when her tether line breaks.
She is able to throw her 12.9 kg oxygen tank away from the shuttle with a speed of 10.4m/s
to propel herself back to the shuttle. Assuming that she starts from rest (relative to the shuttle), determine the maximum distance she can be from the craft when the
line breaks and still return within 52.2 s (the amount of time she can hold her breath). Answer in units of m.

I'm getting the wrong answer to this problem. can someone help me figure out where i went wrong?
This is what I did:
1. Found out what Vf for astronaut would be:
MaVfa= Moxygentank(Vf)
MoVfo/ Ma = Vfa
Vfa= 2.12614897 m/s
2. then i decided to use the distance formula
d= Vi + Vf/2 (t)
where Vi= 0 and Vf= the answer above and t= 52.2s
i got d= 55.49248811m
 
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Your distance formula is for uniformly accelerated motion. This is motion at a constant velocity. How far can you get in 52.2s at a constant velocity of 2.126m/s?
 

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