Answer
$32.9m/s$
Work Step by Step
To reach the maximum speed, the subway can accelerate for half of the distance and decelerate for the second half. This means that the distance over the acceleration is $\frac{806m}{2}=403m$. A formula relating distance, acceleration, initial velocity and displacement is $$v_f^2=v_o^2+2a\Delta x$$ Solving for $v_f$ yields $$v_f=\sqrt{v_o^2+2a\Delta x}$$ Using values of $v_o=0.00m/s$, $a=1.34m/s^2$, and $\Delta x=403m$, the final velocity is equal to $$v_f=\sqrt{(0.00m/s)^2+2(1.34m/s^2)(403m)}$$ $$v_f=32.9m/s$$
