A particle undergoes uniform circular motion on a horizontal xy plane. At time t = 0, it moves through coordinates (3.0 m, 0) with velocity $\overrightarrow{\mathrm{v}}=\left(6.0\mathrm{m}/\mathrm{s}\right)\widehat{\mathrm{j}}$. At t = 5.0 s, it moves through (11.0 m, 0) with velocity $\overrightarrow{\mathrm{v}}=\left(-6.0\mathrm{m}/\mathrm{s}\right)\widehat{\mathrm{j}}$. What is its acceleration at t = 2.5 s?

The motion of the particle is shown in the following figure:

Radius, $\mathrm{r}=\frac{\left(11.0-3.0\right)}{2}=4.0\mathrm{m}.$
Magnitude of acceleration, 
$=\frac{{\mathrm{v}}^2}{\mathrm{r}}=\frac{{\left(6.0\right)}^2}{4.0}=9.0\mathrm{m}/{\mathrm{s}}^2.$
at t = 2.5 s, the particle has completed one-fourth of its period and its velocity is parallel to x-axis. Since the centripetal acceleration is directed towards to the center. 
We get
$\overrightarrow{\mathrm{a}}=\left(-9.0\mathrm{m}/{\mathrm{s}}^2\right)\widehat{\mathrm{j}}$