What happens to a low-mass star after helium flash?
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(a) 3.58 km 45° south of east
The total displacement is given by:
where
v is the average velocity
t is the time
The average velocity is:
v = 3.53 m/s
While we need to convert the time from minutes to seconds:
Therefore, the magnitude of the displacement is
And the direction is the same as the velocity, therefore 45° south of east.
(b) 5.53 m/s 90° south of east
The velocity of the air relative to the ground is
and the direction is exactly opposite to that of Allen, so it is 45° north of west. Allen's velocity relative to the ground is
v = 3.53 m/s
So this must be the resultant of Allen's velocity relative to the air (v') and the air's speed (). Since these two vectors are in opposite direction, we have
Therefore we find v', Allen's velocity relative to the air:
The direction must be measured relative to the air's reference frame. In this reference frame, Allen is moving exactly backward, so his direction will be 90° south of east.
(c) 56.1 km at 90° south of east.
Since Allen's velocity relative to the air is
v' = 5.53 m/s
Then the displacement of Allen relative to the air will be given by
and substituting,
And the direction is the same as that of the velocity, therefore will be 90° south of east.