Answer
$$a =6 \mathrm{m} \cdot \mathrm{s}^{-2}$$
Work Step by Step
To find the acceleration of the cheetah, we need to find the distance that the cheetah would travel and the distance from its prey, and then we must equate them to each other. Thus:
$$
v_{p} t=x
$$
where $v_{c}$ is the velocity of the prey. The cheetah will start from rest (which means $0=v_{0}$ ) with an accelration of $a,$ thus the distance that the cheetah moves is:
$$
\begin{array}{c}
v_{0} t+\frac{1}{2} a t^{2}=x \\
\frac{1}{2} a t^{2}=x
\end{array}
$$
by equating (1) and (2), we get:
$$
\frac{1}{2} a t^{2}=v_{p} t
$$
$$
\frac{2 v_{p}}{t}=a
$$
Plugging in given values:
$$
\begin{aligned}
&\frac{2\left(9 \mathrm{m} \cdot \mathrm{s}^{-1}\right)}{(3 \mathrm{s})}=a \\
&=6 \mathrm{m} \cdot \mathrm{s}^{-1} \\
6 \mathrm{m} \cdot \mathrm{s}^{-2}=a &
\end{aligned}
$$