(a)
In order to find the original speed, let's use the formula below to find an expression for the acceleration:
[tex]\begin{gathered} V=V_0+a\cdot t\\ \\ 2.1=V_0+a\cdot9.45\\ \\ a=\frac{2.1-V_0}{9.45} \end{gathered}[/tex]Now, we can use the following formula to find the initial speed:
[tex]\begin{gathered} \Delta S=V_0t+\frac{at^2}{2}\\ \\ 40=V_0\cdot9.45+\frac{\frac{(2.1-V_0)}{9.45}\cdot9.45^2}{2}\\ \\ 40=9.45V_0+4.725(2.1-V_0)\\ \\ 40=9.45V_0+9.9225-4.725V_0\\ \\ 4.725V_0=40-9.9225\\ \\ V_0=\frac{30.0775}{4.725}\\ \\ V_0=6.3656\text{ m/s} \end{gathered}[/tex](b)
Now, calculating the acceleration, we have:
[tex]\begin{gathered} a=\frac{2.1-6.3656}{9.45}\\ \\ a=-0.4514\text{ m/s^^b2} \end{gathered}[/tex]