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Sagot :
ANSWER
2.5 m/s²
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,
[tex]y-y_0=ut+\frac{1}{2}at^2_{}[/tex]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,
[tex]\Delta y=\frac{1}{2}at^2[/tex]Divide both sides by t² and multiply both sides by 2,
[tex]\begin{gathered} \Delta y\cdot\frac{2}{t^2}=\frac{1}{2}at^2\cdot\frac{2}{t^2} \\ a=\frac{2\Delta y}{t^2} \end{gathered}[/tex]Now replace with the values,
[tex]a=\frac{2\cdot12m}{3.1^2s^2}\approx2.4974m/s^2[/tex]The brick's acceleration is 2.5 m/s², rounded to the nearest tenth.
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