Answer
$v_{max}=22m/s$
Work Step by Step
During a turn, static friction $f_s$ acts as centripetal force $F_c$ that keeps the vehicle along the curve and not skidding off in another direction. In other words, $$f_s=F_c=\frac{mv^2}{r}$$
Since $f_s$ has a maximum value $f_s^{max}$, we have $$f_s^{max}=\frac{mv^2_{max}}{r}$$
The formula for $f_s^{max}$ is $f_s^{max}=\mu_sF_N$. In these cases, since the vehicle has no vertical acceleration, $F_N=mg$, so $f_s^{max}=\mu_smg$
Therefore, $$\mu_smg=\frac{mv^2_{max}}{r}$$ $$\mu_sg=\frac{v^2_{max}}{r}$$
First, we need to find radius of the curve $r$. From car A, we have $\mu_s=1.1$, $v_{max}=25m/s$ and $g=9.8m/s^2$
$$r=\frac{v^2_{max}}{\mu_sg}=58m$$
Now, for car B, we have $\mu_s=0.85$ and $g=9.8m/s^2$
$$v_{max}=\sqrt{\mu_sgr}=22m/s$$
