A spaceship lifts off from the Earth with constant acceleration. 9 seconds into the flight, a small piece of the ship breaks off and falls back to the Earth. If the piece hit the ground 11.5 seconds after breaking off from the spaceship, what is the spaceship’s acceleration (in m/s2)?
To find the spaceship's acceleration, we can use the equations of motion for constant acceleration.
Let's break down the information given:
- The time the piece of the spaceship breaks off and starts falling is 9 seconds.
- The time it takes for the piece to hit the ground after breaking off is 11.5 seconds.
From this, we can determine the time it takes for the piece to fall from the spaceship to the ground:
11.5 seconds - 9 seconds = 2.5 seconds
Now, we can consider the motion of the piece falling to the ground. The relationship between distance (d), initial velocity (v₀), time (t), and acceleration (a) can be described by the equation:
d = v₀t + 0.5at²
In this case, the initial velocity of the piece is 0, because it starts from rest. Therefore, the equation simplifies to:
d = 0.5at²
The distance that the piece falls is the distance between the spaceship and the ground. We can assume the acceleration due to gravity, denoted as "g," is constant and equal to 9.8 m/s².
Substituting the values into the equation:
d = 0.5gt²
Rearranging the equation to solve for acceleration (a):
a = 2d/t²
Since we already calculated that the time it takes for the piece to fall is 2.5 seconds, we can substitute these values into the equation:
a = 2d/2.5²
We are given the distance, d, but to solve for acceleration we need to know the height at which the piece broke off from the spaceship. Without that information, we cannot determine the exact acceleration of the spaceship.