Answer
\begin{array}{l}
3.0 \times 10^{2} \text {days} \\
+1.04 \times 10^{-4} \mathrm{m} / \mathrm{s}^{2}
\end{array}
Work Step by Step
The time $\Delta t$ that it takes for the spacecraft to change its velocity by an amount of $+2700 \mathrm{m} / \mathrm{s}=\Delta v$ is:
$\frac{\Delta v}{a}=\Delta t$
$=>\frac{+2700 \mathrm{m} / \mathrm{s}}{+9.0 \frac{\mathrm{m} / \mathrm{s}}{\mathrm{d} a y}}=\Delta t$
$=>3.0 \times 10^{2}=\Delta t$days
Thus, the acceleration of the spacecraft (in $m / s^{2}$ ) is: $\frac{\Delta v}{t}=a$
$=>\frac{+9.0 \mathrm{m} / \mathrm{s}}{(1 \mathrm{day})\left(\frac{24 \mathrm{hr}}{1 \mathrm{day}}\right)\left(\frac{3600 \mathrm{s}}{1 \mathrm{hr}}\right)}=a$
$=>+1.04 \times 10^{-4} \mathrm{m} / \mathrm{s}^{2}=a$
