Irodov 1.14: Two Lights, One Chord, One Frame – Amit & Deepali Sinha Foundation

Problem Irodov 1.14

A train of length $l = 350$ m starts from rest and moves along a straight track with constant acceleration $w = 3.0\times10^{-2}$ m/s $^2$ . A time $t = 30$ s after the start, the headlight at the front of the train switches on (event 1); a further time $\tau = 60$ s later, the tail light at the back of the train switches on (event 2). Find the spatial separation between events 1 and 2 as measured in a frame fixed to the train and as measured in a frame fixed to the ground. Then find the constant velocity $V$ , relative to the ground, of a frame $K$ in which the two events occur at the same spatial point, and state its direction.

Position versus time for the front (locomotive) of the train, x_F(t), and the back (tail), x_B(t), offset by l. Event 1 sits on the front's curve at t=30 s, event 2 on the back's curve at t=90 s. — Position versus time for the front (locomotive) of the train, $x_F(t)$ , and the back (tail), $x_B(t)$ , offset by $l$ . Event 1 sits on the front’s curve at $t=30$ s, event 2 on the back’s curve at $t=90$ s.

Given

$l = 350$ m
$w = 3.0\times10^{-2}$ m/s $^2$
$t = 30$ s
$\tau = 60$ s

Find

distance between events 1 and 2 in the train frame and in the Earth frame; the velocity

V

of frame

K

Solution

Step 1Position of each end of the train

The train starts from rest at $t=0$ and accelerates rigidly, so every point of it shares the same velocity and displacement history; only the two ends’ starting positions differ, by $l$ . Taking the front’s starting point as the origin, with $x$ positive in the direction of motion, and writing $x_F(t)$ for the front (locomotive) and $x_B(t)$ for the back (tail),

\begin{align*} x_F(t) &= \frac{1}{2}w t^2 \tag{1} \\ x_B(t) &= -l+\frac{1}{2}w t^2 \tag{2}\end{align*}

give the front’s and the back’s position at any time $t$ ; the front leads, so it is the larger, rightmost coordinate throughout.

Step 2Distance between the events in the train frame

Event 1 occurs at the front and event 2 at the back, and both points are fixed in the frame attached to the train, a distance $l$ apart along the train regardless of when each event happens.

\Delta x_{\text{train}} = l

(3)

Step 3Distance between the events in the Earth frame

Event 1 happens at time $t_1=t$ at the front, event 2 at time $t_2=t+\tau$ at the back. Substituting these times into (1) and (2) and subtracting gives their Earth-frame separation.

\begin{align*}x_1-x_2 &= \frac{1}{2}w t_1^2 -\left(-l+\frac{1}{2}w t_2^2\right) \\ &= l-\frac{1}{2}w\left(t_2^2-t_1^2\right) \\ &= l-\frac{1}{2}w(t_2-t_1)(t_2+t_1)\end{align*}

The two elapsed times are $t_2-t_1=\tau$ and $t_2+t_1=2t+\tau$ , so the product becomes $\tau(2t+\tau)$ ; halving that bracket puts the separation in closed symbolic form.

x_1-x_2 = l-\frac{1}{2}w\,\tau(2t+\tau) = l-w\tau\!\left(t+\frac{\tau}{2}\right) \tag{4}

Substituting $l=350$ m, $w=3.0\times10^{-2}$ m/s $^2$ , $t=30$ s and $\tau=60$ s,

x_1-x_2 = 350-(3.0\times10^{-2})(60)\!\left(30+30\right)\ \text{m} = 350-108\ \text{m}

\begin{aligned}\Delta x_{\text{Earth}} &= l-w\tau\left(t+\dfrac{\tau}{2}\right)\\ &= 2.42\times10^{2}\text{ m}\approx0.24\text{ km}\end{aligned}

(5)

The positive sign means event 1 sits ahead of event 2 along the direction of motion.

Step 4The velocity of frame

K

The same pair of worldlines with the chord joining event 1 to event 2. A frame whose own worldline is that chord passes through both events, so its constant velocity is the chord's slope; this step computes it. — The same pair of worldlines with the chord joining event 1 to event 2. A frame whose own worldline is that chord passes through both events, so its constant velocity is the chord’s slope; this step computes it.

A point fixed in $K$ and located at $x_1$ at time $t_1$ is at $x_1+V(t_2-t_1)$ at time $t_2$ . For the two events to coincide in $K$ , this must equal $x_2$ .

\begin{align*}x_1+V(t_2-t_1) &= x_2 \\ V &= \dfrac{x_2-x_1}{t_2-t_1} \\ &= -\dfrac{x_1-x_2}{\tau}\end{align*}

Substituting (4) for $x_1-x_2$ ,

V = -\frac{1}{\tau}\left[\,l-w\tau\!\left(t+\frac{\tau}{2}\right)\right].

Distributing the $-1/\tau$ across the bracket leaves $V$ in terms of the given quantities alone.

V = w\!\left(t+\frac{\tau}{2}\right)-\frac{l}{\tau}

Evaluating each term with the given numbers,

V = (3.0\times10^{-2})(60)-\frac{350}{60}\ \text{m/s} = 1.8-5.8\ \text{m/s}

V = w\!\left(t+\dfrac{\tau}{2}\right)-\dfrac{l}{\tau} = -4.0\text{ m/s}

(6)

The minus sign means $K$ moves opposite to the train’s direction of travel, at $4.0$ m/s.

Check

Dimensionally each boxed result is a length or a length over time, as required. As a limiting check, if $\tau\to0$ the two events become simultaneous and Step 4’s chord slope diverges, $V\to-\infty$ : no finite-velocity frame can bring two simultaneous, spatially separated events to the same point, which is exactly the sign that the construction breaks down there.

Remark

The chord in the figure is the same idea as the mean-velocity chord of a single moving point (compare Irodov 1.28): here the two endpoints belong to different physical points (front, then back) rather than to one trajectory, so the chord’s slope is not any material particle’s velocity, only the velocity a bookkeeping frame must have to file both events under one address.