given that acceleration due to gravity is g = 10 m/s^2
speed of the rocket will reach to 0.9c
[tex]v_f = 0.9* 3 * 10^8 = 2.7 * 10^8 m/s[/tex]
now by kinematics
[tex]v_f = v_i + a*t[/tex]
[tex]2.7 * 10^8 = 0 + 10*t[/tex]
[tex]t = 2.7*10^7 s[/tex]
Part b)
distance traveled by it in above time
[tex]d = v*t + \frac{1}{2}at^2[/tex]
[tex]d = 0 + \frac{1}{2}*10*(2.7*10^7)^2[/tex]
[tex]d = 3.645 * 10^{15} m[/tex]
so this is the distance covered by the object