Answer
$\Delta t = 49.0s$
Work Step by Step
Since the magnitude of the acceleration is the same over the entire ride, since the subway accelerates through 403m and decelerates through 403m, the amount of time to get from the start to 403m is the same as the time from 403m to 806m. Therefore, using known values of $\Delta x=403m$, $a=1.34m/s^2$, and $v_o=0.00m/s^2$, a kinematics equation can be used. This is $$\Delta x = \frac{1}{2}a\Delta t^2+v_o\Delta t$$ Since $v_o=0.00m/s$, the equation becomes $$\Delta x =\frac{1}{2}a\Delta t^2$$ Solving for $\Delta t$ yields $$\Delta t = \sqrt{\frac{2\Delta x}{a}}$$ Using known values of $\Delta x=403m$ and $a=1.34m/s^2$ yields a change in time of $$\Delta t = \sqrt{\frac{2(403m)}{1.34m/s^2}}=24.5s$$ This is only half the trip. To cover the whole trip, multiply the time by two to get get $2(24.5s)=49.0s$.
