Question

a rocket is launched straight up with constant acceleration. 2.8 seconds after liftoff, a bolt falls off the side of the rocket. the bolt hits the ground 9.2 seconds later. what was the rocket's acceleration (in m/s²)?

Answer 1

Using the kinematic equation, the rocket's acceleration is calculated by setting up and solving an equation based on the displacement of a bolt that falls off the rocket.

To determine the rocket's acceleration, we can use the kinematic equation that relates displacement, initial velocity, time, and acceleration:

`\[ s = ut + (1)/(2)at^2 \]`

Where:

( s ) is the displacement (in this case, the distance the bolt falls, which is the same as the rocket's upward distance).

• u is the initial velocity.

• a is the acceleration.

• t is the time.

Let's denote the upward direction as positive. Initially, the bolt has the same velocity as the rocket, so u = 0 (initially at rest relative to the rocket).

After 2.8 seconds, the displacement of the bolt is given by
`\( s = (1)/(2) a (2.8)^2 \)`.

The total time the bolt is in motion is 9.2 seconds, so the displacement during this time is
`\( s = ut + (1)/(2)at^2 \)` with u = 0 and t = 9.2 - 2.8 .

Now, we can set up an equation using the information provided:

`\[ (1)/(2) a (2.8)^2 + (1)/(2) a (6.4)^2 = (1)/(2) a (9.2)^2 \]`

Solving for a , we find the rocket's acceleration. The calculation involves squaring, multiplying, and rearranging terms.

The solution to the equation yields the acceleration of the rocket.