Answer:
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
Explanation:
Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ([tex]a[/tex]), measured in meters per square second, is estimated by this kinematic formula:
[tex]a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }[/tex] (1)
Where:
[tex]\Delta s[/tex] - Travelled distance, measured in meters.
[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speeds of the spaceship, measured in meters.
If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 10968\,\frac{m}{s}[/tex] and [tex]\Delta s = 212.8\,m[/tex], then the acceleration experimented by the spaceship is:
[tex]a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}[/tex]
[tex]a = 282652.782\,\frac{m}{s^{2}}[/tex]
The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.
