Astronaut's Maximum Distance from Shuttle Due to Line Breakage

[Maiia]

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

 

[Dick]

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?

 

[thomate1]

Your approach is correct, but your calculation for Vf is incorrect. You used the mass of the oxygen tank instead of the astronaut in the equation. The correct calculation for Vf would be:

MaVfa = Mava
(63.1 kg)(Vfa) = (12.9 kg)(10.4 m/s)
Vfa = 2.12614897 m/s

Using this value for Vf, the maximum distance the astronaut can be from the shuttle and still return within 52.2 s would be:

d= Vi + Vf/2 (t)
d= 0 + (2.12614897 m/s)/2 (52.2 s)
d= 55.49248811 m

So your final answer is correct. It seems like you just made a small calculation error in finding Vf. Always double check your calculations to ensure accuracy.