Question

An astronaut standing on the Mars drops a brick. If the brick falls 12m vertically in 3.1s,what is its acceleration?

Answer 1

2.5 m/s²

Step-by-step explanation

We have to find the acceleration given the brick's vertical displacement and the time until it reaches the ground.

The vertical displacement of an object in free fall is,

`y-y_0=ut+(1)/(2)at^2_{}`

y - y0 = Δy is the displacement, 12m. u is the initial velocity, which in this case is 0 becase the astronaut drops the brick. And t is the time 3.1s.

Before replacing with the values, let's solve for a,

`\Delta y=(1)/(2)at^2`

Divide both sides by t² and multiply both sides by 2,

`\begin{gathered} \Delta y\cdot(2)/(t^2)=(1)/(2)at^2\cdot(2)/(t^2) \\ a=(2\Delta y)/(t^2) \end{gathered}`

Now replace with the values,

`a=(2\cdot12m)/(3.1^2s^2)\approx2.4974m/s^2`

The brick's acceleration is 2.5 m/s², rounded to the nearest tenth.