Question

In a novel from 1866 the author describes a spaceship that is blasted out of a cannon with a speed of about 1.69×104 m/s. The spaceship is approximately 274 m long, but part of it is packed with gunpowder, so it accelerates over a distance of only 203m.

What was the acceleration (in m/s2) experienced by the occupants of the spaceship during launch? (Though the novel’s author realized that the “travelers would … encounter a violent recoil,” he probably didn’t know that people generally lose consciousness if they experience accelerations greater than about 7g∼70m/s2.)

Answer 1

To find the acceleration experienced by the occupants of the spaceship during launch, we need to use the kinematic equation:

v^2 = u^2 + 2as

where v is the final velocity (1.69×10^4 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the distance over which the acceleration occurs (203 m).

Rearranging the equation, we get:

a = (v^2 - u^2) / 2s

Plugging in the values, we get:

a = (1.69×10^4 m/s)^2 / (2 × 203 m) ≈ 1.42×10^7 m/s^2

This is a very high acceleration, much greater than the limit of 70 m/s^2 that people can withstand without losing consciousness. Therefore, the occupants of the spaceship would likely lose consciousness during launch.

Answer 2

We can use the kinematic equation:

v^2 = u^2 + 2as

where v is the final velocity (1.69×104 m/s), u is the initial velocity (0 m/s), a is the acceleration, and s is the distance over which the acceleration occurred (203 m).

Solving for a, we get:

a = (v^2 - u^2) / (2s)

a = (1.69×104 m/s)^2 / (2 × 203 m)

a = 1.4074×10^6 m/s^2

This is the acceleration experienced by the spaceship occupants during the launch. It is an extremely high acceleration, over 200 times the acceleration due to gravity (9.8 m/s^2), and far beyond the limit of what humans can tolerate without losing consciousness.