## Physics (10th Edition)

The driver must reduce the speed to $12.1m/s$.
In an unbanked turn, $f_s$ provides the centripetal force that keeps the car safely turning. In formula, $$f_s=F_c=\frac{mv^2}{r}$$ $$v^2=\frac{f_sr}{m} (1)$$ The maximum static friction on wet-road $f_w^{max}$ is $1/3$ that on dry road $f_d^{max}$. As we know the car's speed on dry road $v_d=21m/s$, from equation (1), we have $$\frac{v_w^2}{v_d^2}=\frac{\frac{f_w^{max}r}{m}}{\frac{f_d^{max}r}{m}}=\frac{f_w^{max}}{f_d^{max}}=\frac{1}{3}$$ $$v_w^{max}=\frac{v_d}{\sqrt3}=12.1m/s$$ So the driver must reduce the speed to $12.1m/s$.