We can use the third equation of motion, which is $v_f^2 = v_i^2 + 2as$. Here, $v_f$ is the final speed, $v_i$ is the initial speed, $a$ is the acceleration, and $s$ is the distance. Substituting the given values, we get:
\[2.59^2 = 0^2 + 2a(1.31)\]
Solving for
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