When acceleration is constant, the average velocity is given by
[tex]\bar v=\dfrac{v+v_0}2[/tex]
where [tex]v[/tex] and [tex]v_0[/tex] are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by
[tex]\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}[/tex]
where [tex]x,x_0[/tex] are the final/initial displacements, and [tex]t,t_0[/tex] are the final/initial times, respectively.
Take the car's starting position to be at [tex]t_0=0\,\mathrm s[/tex]. Then
[tex]\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t[/tex]
So we have
[tex]x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m[/tex]
You also could have first found the acceleration using the equation
[tex]v=v_0+at[/tex]
then solve for [tex]x[/tex] via
[tex]x=x_0+v_0t+\dfrac12at^2[/tex]
but that would have involved a bit more work, and it turns out we didn't need to know the precise value of [tex]a[/tex] anyway.
