Solution

We know the instantaneous velocity and acceleration of any object at a given point can be extracted from a plot using
\begin{align*} v &= \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{d x}{d t} \end{align*}
\begin{align*} a &= \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} =\frac{d v }{d t} \end{align*}
We note that, for any given time, the graphs exactly coincide, so any $\Delta x$ is exactly the same for both particles, and so $dx_1 = dx_2$. Likewise, for any time interval, we know that $\Delta t_1 = 4 \Delta t_2$, so $dt_1 = 4dt_2$. We can then say, for an equivalent Point A on their graphs, that:
\begin{align*} v_1 &=\frac{d x_1}{d t_1} = \frac{d x_2}{4 d t_2} = \frac{1}{4} \left( \frac{d x_2}{d t_2} \right) = \frac{1}{4} v_2 \end{align*}
More generally, we can say that, if the time scale divisions are related by $\Delta t_1 / \Delta t_2 = \alpha$, the ratio of velocities are given by:
\begin{align*} \frac{v_2}{v_1} = \alpha \end{align*}
Using the same logic, we can then say for the accelerations that:
\begin{align*} a_1 &= \frac{d v_1 }{d t_1} = \frac{d \left( \frac{1}{4} v_2 \right)}{4 d t_2} = \frac{1}{16} \left( \frac{d v_2}{d t_2} \right) = \frac{1}{16} a_2 \end{align*}
More generally, the ratio of accelerations is given by:
\begin{align*} \frac{a_2}{a_1} = \alpha^2 \end{align*}