Answer:
The value is [tex]v_a = 0.489 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the spacesuit plus astronaut is [tex]m = 191 \ kg[/tex]
The average speed of air molecules is [tex]v = 549 \ m/s[/tex]
The density of air molecule is [tex]\rho = 1.05 \ kg / m^3[/tex]
The acceleration of the air from Whatney's suit is [tex]a = 0.020 \ m/s^2[/tex]
The time considered is [tex]t = 8.1 \ s[/tex]
Generally the mass of air that have left Whatney's suit after the time considered is mathematically represented as
[tex]M = a * \rho * t[/tex]
=> [tex]M = 0020 * 1.05 * 8.1[/tex]
=> [tex]M = 0.1701 \ kg[/tex]
Generally the momentum of the escaped air is
[tex]p = M * v[/tex]
=> [tex]p = 0.1710 * 549[/tex]
=> [tex]p = 93.38 \ kg \cdot m/s[/tex]
Generally from the law of momentum conservation
[tex]p = p_a[/tex]
Here [tex]p_a[/tex] is the momentum of the astronaut at the considered time
[tex]93.38 = m * v_a[/tex]
Here [tex]v_a[/tex] is the velocity of the astronaut at the considered time
=> [tex]93.38 = 191 * v_a[/tex]
=> [tex]v_a = 0.489 \ m/s[/tex]
