let's assume that the distance across the solar system in meters is given by D= 1.50 X 10^11 m.
a) The diagram showing the motion speed as a function of distance is attached below. For the first half of the trip the slope of the speed is positive (it means positive acceleration) for the last half of the trip the slope of the speed is negative. The maximum speed is `c/2`.
b) The initial acceleration direction is positive (toward positive values of x axis). Considering non-relativistic motion the value of the acceleration comes from
`c^2 =4a*D rArr a =c^2/(4D) =(3*10^8)^2/(4*1.5*10^11) =1.5*10^5 m/s^2`
c) the acceleration value on the last half of the trip is equal to the acceleration computed above except a minus sign in front of it. This comes from symmetry considerations.
`a =-1.5*10^5 m/s^2`
d) Assuming non-relativistic effects the time to accelerate to half the speed of light (at half trip) is
`c/2 =a*t_1 rArr t_1 =c/(2a) =3*10^8/1.5*10^5 =2000 s`
since the deceleration on the last half of the trip is equal to the acceleration the total time is
`t = 2*t_1 =4000 s =66.67 min =1h 6.67 min =1h6min40 sec`
To take into account the relativistic effects, the time slows for the moving observer. The average velocity of the ship is
`V =(c/2)/2 =c/4 =0.25*c`
The time for an observer in the ship is
`t' =t/sqrt(1-v^2/c^2) =4000/sqrt(1-0.25^2) =4131.2 sec =68.85 min`
The thrust force is
`F =M*a =2.85*10^8 *1.5*10^5 =4.275*10^13 N`
The number of `g` is simply
`N = (a/g) = (1.5*10^5)/9.8 =15306`
Since the maximum sustainable g for a human is only 3, a person would be squashed the acceleration of the ship in its trip.