Assuming the acceleration stays constant, we have
[tex]a=\dfrac{v-v_0}{t-t_0}[/tex]
where [tex]v[/tex] is the velocity of the car at time [tex]t[/tex], [tex]v_0=0[/tex] is the starting velocity, and [tex]t_0=0[/tex] is the starting time. So
[tex]a=\dfrac vt\iff7.815\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{37.0\,\frac{\mathrm m}{\mathrm s}}t\implies t=\dfrac{37.0\,\frac{\mathrm m}{\mathrm s}}{7.815\,\frac{\mathrm m}{\mathrm s^2}}[/tex]
[tex]\implies t\approx4.73\mathrm s[/tex]