Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So
[tex]a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]
Now, we have that
[tex]{v_f}^2-{v_0}^2=2a\Delta x[/tex]
so we end up with a distance traveled of
[tex]\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x[/tex]
[tex]\implies\Delta x=79.6\,\mathrm m[/tex]
