Source: Principles of Physics
Problem Sets:
Problem
A subway train travels a distance $d$ between two stations. The train starts from rest at the first station and comes to a complete stop at the second. The maximum tolerable acceleration magnitude for passengers is $a$. The train stops for a duration $t_{stop}$ at each station.
- What is the maximum speed $v_{max}$ the train can attain?
- What is the travel time $t_{travel}$ between stations?
- What is the maximum average speed $v_{avg}$ of the train, from one start-up to the next?
- Describe the graphs of position $x$, velocity $v$, and acceleration $a$ versus time $t$.
[Q1] $v_{max} = \sqrt{ad}$ [Q2] $t_{travel} = 2\sqrt{\frac{d}{a}}$ [Q3] $v_{avg} = \frac{d}{2\sqrt{d/a} + t_{stop}}$ [Q4] The graphs are described in the solution.
To minimize travel time and maximize speed, the train accelerates with $+a$ for the first half of the distance, $d/2$, and then decelerates with $-a$ for the second half.
[Q1] We use the time-independent kinematic equation $v_f^2 = v_i^2 + 2a\Delta x$ for the first half of the trip ($v_i=0$, $\Delta x = d/2$).
$$v_{max}^2 = 0 + 2a\left(\frac{d}{2}\right) = ad$$[Q2] The time to accelerate to $v_{max}$ is $t_1$. From $v_f = v_i + at$:
$$t_1 = \frac{v_{max}}{a} = \frac{\sqrt{ad}}{a} = \sqrt{\frac{d}{a}}$$By symmetry, the deceleration time $t_2$ is equal to $t_1$. The total travel time is the sum of the two.
$$t_{travel} = t_1 + t_2 = 2t_1$$[Q3] The average speed is the total distance traveled during one cycle divided by the total time for that cycle. A cycle includes travel time and stop time.
$$v_{avg} = \frac{d}{t_{travel} + t_{stop}}$$[Q4] The motion graphs cover the full cycle time $T = t_{travel} + t_{stop}$.
- Acceleration ($a$ vs $t$): A step function. It is $+a$ for $0 < t < t_1$, then $-a$ for $t_1 < t < t_{travel}$, and $0$ for $t_{travel} < t < T$.
- Velocity ($v$ vs $t$): A triangle shape. It increases linearly from $0$ to $v_{max}$ until $t=t_1$, decreases linearly back to $0$ until $t=t_{travel}$, and remains $0$ until $t=T$.
- Position ($x$ vs $t$): Two connected parabolic segments. The position increases as a parabola opening upward until $x=d/2$ at $t=t_1$, then as a parabola opening downward until $x=d$ at $t=t_{travel}$. The position remains constant at $x=d$ until $t=T$.