Answer
Ans:
$$
a_P=774 \mathrm{ft} / \mathrm{s}^2
$$
Work Step by Step
First we need to determine the angular velocity of gear $A$ at $t=0.75 \mathrm{~s}$.
For that we use the following equation.
$$
\begin{gathered}
d \omega=\alpha d t \\
\int_0^{\omega_A} d \omega=\int_0^{0.75 \mathrm{~s}} 300 \sqrt{t} d t \\
\omega_A=\left.200 t^{3 / 2}\right|_0 ^{0.75 \mathrm{~s}}=129.9 \mathrm{rad} / \mathrm{s} \\
\alpha=300 \sqrt{t}
\end{gathered}
$$
So the angular acceleration of gear $A$ is:
$$
\alpha_A=300 \sqrt{0.75}=259.81 \mathrm{rad} / \mathrm{s}^2
$$
$$
\begin{gathered}
\omega_A r_A=\omega_B r_B \quad \text { and } \quad a_A r_A=a_B r_B \\
\omega_B=\frac{r_A}{r_B} \omega_A=\left(\frac{0.7}{1.4}\right)(129.9)=64.95 \mathrm{rad} / \mathrm{s} \\
\alpha_B=\frac{r_A}{r_B} \alpha_A=\left(\frac{0.7}{1.4}\right)(259.81)=129.9 \mathrm{rad} / \mathrm{s}^2
\end{gathered}
$$
$\omega_B=\omega_p$ and $\alpha_B=\alpha_p$; where $P_{\text {refers to }}$ the propeller.
$$
\begin{gathered}
v_P=\omega_B r_P=64.95\left(\frac{2.20}{12}\right)=11.9 \mathrm{ft} / \mathrm{s} \\
a_t=\alpha_B r_P=129.9\left(\frac{2.20}{12}\right)=23.82 \mathrm{ft} / \mathrm{s}^2 \\
a_n=\omega_B^2 r_P=(64.95)^2\left(\frac{2.20}{12}\right)=773.44 \mathrm{ft} / \mathrm{s}^2
\end{gathered}
$$
So the magnitude of the acceleration of $P$ is:
$$
a_p=\sqrt{a_t^2+a_n^2}=\sqrt{(23.82)^2+(773.44)^2}=774 \mathrm{ft} / \mathrm{s}^2
$$
