Physics · Mechanics · Grade 7

Metro: from station to station

How does a metro train stop in exactly the right place at the next station?

The situation

On the Tashkent Metro a train leaves a station: first it accelerates, then moves at constant speed, then brakes and stops at the next station exactly opposite the stopping mark. The hand straps in the carriage swing now backwards, now forwards. Where must the driver begin braking, and what is the average speed between stations?

Topic

Uniform and non-uniform motion. Tashkent Metro: a speed graph between stations

Key ideas

Uniformly accelerated motion
Average and instantaneous speed
Area under the v(t) graph

Formulas

v = v_{0} + a t, s = v_{0} t + \frac{a t^{2}}{2}

Uniformly accelerated motion: speed changes linearly, and distance changes with the square of time.

s_{1} = \frac{v^{2}}{2 a_{1}}, s_{3} = \frac{v^{2}}{2 a_{2}}, x_{t} = D - \frac{v^{2}}{2 a_{2}}

Acceleration and braking distances; $x_{t}$ is the point where braking begins: the train stops exactly at the distance $D$ .

v = \frac{d x}{d t}, a = \frac{d v}{d t}, x = \int_{0}^{t} v d t, a = \frac{2 s}{t^{2}}

Speed is the slope of the tangent to $x (t)$ , distance is the area under $v (t)$ . The last is the formula for measuring acceleration in uniformly accelerated motion from rest.

v_{o ‘ r} = \frac{s}{t} = \frac{D}{t_{1} + t_{2} + t_{3}}

Average speed is the whole distance divided by the whole time. It equals the area under the $v (t)$ graph divided by the time.

\tan φ = \frac{a}{g}

The angle of the hand strap's swing shows the acceleration (an accelerometer).

Theory

Uniform and non-uniform motion

The train travels between stations in three stages. During acceleration the speed increases by $a_{1}$ every second, and the distance graph $x (t)$ is a parabola. In uniform motion the speed is constant, $x (t)$ is a straight line, $a = 0$ . During braking the speed decreases steadily. Sitting in the carriage during uniform motion, you do not feel that you are moving — the straps hang vertically.

Instantaneous and average speed

The speedometer shows the instantaneous speed. The average speed $v_{a v g} = s / t$ is the ratio of the whole distance to the whole time. Because of acceleration and braking it is always less than $v_{m a x}$ . The area under the $v (t)$ graph equals the distance covered: the area of the trapezium $\frac{1}{2} v_{m a x} (T + t_{2})$ gives exactly $D$ .

Where should the brake be applied?

A train with speed $v$ stops with deceleration $a_{2}$ over a distance $s_{3} = v^{2} / 2 a_{2}$ . So braking must begin at the point $x_{t} = D - s_{3}$ . If $s_{1} + s_{3} > D$ , the train has no time to reach $v_{m a x}$ : the speed graph becomes a triangle, and the greatest speed is $v_{p} = \sqrt{2 D a_{1} a_{2} / (a_{1} + a_{2})}$ .

The strap is an accelerometer

In an accelerating carriage it is as if a force $- m a$ , directed backwards because of inertia, acts on the strap. In equilibrium the string is stretched along the sum of the weight and this force: $\tan φ = a / g$ . During acceleration the strap swings back, during braking forward, and then it oscillates a little. Model: the accelerations are constant and switch instantly; gradient, bends and air resistance are neglected.

Model, frame of reference and types of motion

We take the 59 m long train as a material point (particle): all its points move in the same way (translational motion), and a single $x$ gives its position. The carriages do not deform — an absolutely rigid body; the three carriages are a system of particles; the air in the tunnel is a continuous medium. The train is made of matter, and is pulled down by the Earth's gravitational field (two forms of matter); “empty” space without particles — the physical vacuum — also has its own properties in quantum physics. Motion is relative: relative to the platform a passenger moves with the train, relative to the carriage he is at rest (the “Inside the carriage” camera). A frame of reference = a reference body + coordinates + a clock. The trajectory here is a straight line, so the distance travelled equals the magnitude of the displacement; if the train went back, the distance would be $2 D$ and the displacement zero. The wheels rotate, while the train moves translationally. If the tunnel bends, a normal acceleration $a_{n} = v^{2} / R$ appears even at constant speed, and the straps swing sideways — curvilinear motion is in the “Banked turn” and “Circular motion: the bicycle” experiments.

Graphs: derivative, area and scale

Position, velocity and acceleration are vector quantities; on a straight line the sign shows their direction ( $a < 0$ when braking). Instantaneous speed is the slope of the tangent to the $x (t)$ graph, $v = d x / d t$ (the derivative); in uniform motion the graph is a straight line, in acceleration a parabola. Acceleration $a = d v / d t$ is the slope of $v (t)$ . The inverse operation is the integral: distance is the area under the $v (t)$ graph. To turn area into metres a scale is needed: a rectangle of 10 s × 10 km/h on the graph equals $10 \cdot 10 / 3.6 \approx 27.8$ m of distance. Engineers design train timetables and braking curves on such graphs. Lab work: when the train starts from rest, the time at which its front passes each 25 m mark is entered in a table; from $s = a t^{2} / 2$ , $a = 2 s / t^{2}$ . Acceleration can also be found from the strap's deflection: $a = g \tan φ$ .

Did you know?

The Tashkent Metro opened in 1977 — it was the first metro in Central Asia.
Acceleration can be measured from the strap's deflection: at $a = 1$ m/s², $φ = \arctan (1 / 9.81) \approx {5.8}^{\circ}$ . The accelerometer in smartphones is also based on the phenomenon of inertia.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

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