Answer
$v_B=22m/s$
Work Step by Step
The formula for a banked curve is $$\tan\theta=\frac{v^2}{rg} (1)$$
We have $r_A=r_B=r$, $\theta_A=13^o$, $\theta_B=19^o$, $v_A=18m/s$
To find $v_B$, we rely on formula (1): $$\frac{\tan\theta_A}{\tan\theta_B}=\frac{\frac{v_A^2}{rg}}{\frac{v_B^2}{rg}}=\frac{v_A^2}{v_B^2}$$ $$\frac{v_A^2}{v_B^2}=\frac{\tan13^o}{\tan19^o}=0.67$$ $$v_B=\frac{v_A}{\sqrt{0.67}}=22m/s$$
