Let the mutual force between the two astronauts when they push apart be
F N, as shown in the figure below.
1st astronaut:
The mass is 60 kg.
Let the acceleration be a₁ m/s².
Then
F = 60a₁ (1)
2nd astronaut:
The mass is 90kg.
The acceleration, a₂, is given by
F = 90a₂ (2)
When lighter astronaut has traveled a distance of 15 m, the time taken, t, is given by
(1/2)a₁t² = 15
or
a₁ t² = 30 (3)
The distance traveled by the other astronaut is
d₂ = (1/2)a₂ t² (4)
Substitute (3) into (4)
d₂ = (1/2) a₂ (30/a₁) = 15 (a₂/a₁)
The separation between the astronauts is
D = 15 + d₂ = 15 + 15(a₂/a₁) (5)
From (1) and (2), obtain
60a₁ = 90a₂
a₂/a₁ = 2/3 (6)
Substitute (6) into (5).
D = 15 + 15(2/3) = 25 m
Answer: 25 m
