Let [tex]a_1[/tex] be the average acceleration over the first 2.46 seconds, and [tex]a_2[/tex] the average acceleration over the next 6.79 seconds.
At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let [tex]v[/tex] be the velocity of the car after the first 2.46 seconds.
By definition of average acceleration, we have
[tex]a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}[/tex]
[tex]a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}[/tex]
and we're also told that
[tex]\dfrac{a_1}{a_2}=1.66[/tex]
(or possibly the other way around; I'll consider that case later). We can solve for [tex]a_1[/tex] in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:
[tex]1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}[/tex]
[tex]\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}[/tex]
Now we can solve for [tex]v[/tex]. We find that
[tex]v=20.8\,\dfrac{\mathrm m}{\mathrm s}[/tex]
In the case that the ratio of accelerations is actually
[tex]\dfrac{a_2}{a_1}=1.66[/tex]
we would instead have
[tex]\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)[/tex]
in which case we would get a velocity of
[tex]v=24.4\,\dfrac{\mathrm m}{\mathrm s}[/tex]
