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Sagot :
Since the object is moving with uniform acceleration we have an uniformly accelerated motion which means that we can use the following equations:
[tex]\begin{gathered} a=\frac{v_f-v_0}{t} \\ x=x_0+v_0t+\frac{1}{2}at^2 \\ v_f^2-v_0^2=2a(x-x_0) \end{gathered}[/tex]Now, in this case we know:
• The initial position 2.76 cm.
,• The initial velocity 13 cm/s
,• The final position -5 cm
,• The time it takes 3.05 s.
And we want to determine the acceleration; from what we know and what we want we determine that we can use the second equation. Plugging the values in that equation we have that:
[tex]\begin{gathered} -5=2.76+(13)(3.05)+\frac{1}{2}(3.05)^2a \\ \frac{3.05^2}{2}a=-5-2.76-(13)(3.05) \\ \frac{3.05^2}{2}a=-47.41 \\ 3.05^2a=-94.82 \\ a=-\frac{94.82}{3.05^2} \\ a=-10.19 \end{gathered}[/tex]Therefore, the acceleration is -10.19 cm/s²
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