Answer:
[tex]a\approx 0.56 \,\frac{m}{s^2}[/tex]
[tex]v_f=6.72\,\frac{m}{s}[/tex]
Explanation:
This is a motion under constant acceleration, so we can use the kinematic equation for the distance covered:
[tex]x_f-x_i=v_i\,t+\frac{1}{2} a\,t^2\\40=0+\frac{1}{2} a\,(12)^2\\40=72 \,a\\a=\frac{40}{72} \,\frac{m}{s^2} \\a\approx 0.56 \,\frac{m}{s^2}[/tex]
Now, the final velocity can be calculated via:
[tex]v_f=v_i+a\,t\\v_f=0+0.56\,(12)\\v_f=6.72\,\frac{m}{s}[/tex]
