Respuesta :
[tex]8.10\frac{m}{s^2}[/tex]
Explanation
to solve this we need to use the expression
[tex]\begin{gathered} x=v_it+\frac{1}{2}at^2 \\ \text{where} \\ vf\text{ is the final sp}eed \\ vi\text{ is the initial spe}ed \\ a\text{ is the acceleration} \\ x\text{ is the traveled distance} \end{gathered}[/tex]Step 1
Let
[tex]\begin{gathered} vi=0(\text{ it starts from the rest)} \\ a=a \\ vt=5.21\text{ s} \\ x=110\text{ m} \end{gathered}[/tex]replace
[tex]\begin{gathered} x=v_it+\frac{1}{2}at^2 \\ 110m=(0)(2.51s)+\frac{1}{2}(a)(5.21s)^2 \\ 110m=0+\frac{1}{2}(a)(5.21s)^2 \\ 110m=13.57s^2\cdot a \\ \text{divide both sides by }13.57s^2 \\ \frac{110m}{13.57s^2}=\frac{13.57s^2\cdot a}{13.57s^2} \\ 8.10\frac{m}{s^2}=a \end{gathered}[/tex]therefore, the answer is
[tex]8.10\frac{m}{s^2}[/tex]I hope this helps you
