Irodov 1.13: Aim Steadily at a Target That Cannot Turn

Problem Irodov 1.13

Point $A$ moves at a constant speed $v$ , its velocity vector always pointed at the current position of point $B$ . Point $B$ moves in a straight line at a constant speed $u$ , with $u<v$ . At $t=0$ the two points are a distance $l$ apart and $A$ ’s velocity is perpendicular to $B$ ’s velocity $\vect u$ . Find the time $\tau$ for $A$ to reach $B$ .

The initial instant. A and B are a distance l apart; B travels along the straight line drawn through it at velocity u, and A's initial velocity v_0 is aimed at B, perpendicular to u. — The initial instant. $A$ and $B$ are a distance $l$ apart; $B$ travels along the straight line drawn through it at velocity $\vect u$ , and $A$ ’s initial velocity $\vect v_0$ is aimed at $B$ , perpendicular to $\vect u$ .

Given

$v$
$u<v$
initial separation $l$
$\vect v_0\perp\vect u$

Find

the time

\tau

for

A

to reach

B

Solution

Step 1The closing rate along the line

AB

The chase. A curves inward, always aimed at B's current position, while B runs up its straight line; they meet at M. At a later instant α is the angle between A's velocity v and the dashed reference parallel to u. — The chase. $A$ curves inward, always aimed at $B$ ’s current position, while $B$ runs up its straight line; they meet at $M$ . At a later instant $\alpha$ is the angle between $A$ ’s velocity $\vect v$ and the dashed reference parallel to $\vect u$ .

Let $r$ be the instantaneous separation and $\alpha$ the angle between the line $AB$ (equivalently, $A$ ’s own velocity $\vect v$ , since it is always aimed at $B$ ) and $B$ ’s velocity $\vect u$ . Since $A$ moves directly toward $B$ at speed $v$ , and $u$ ’s component along $AB$ is $u\cos\alpha$ , the rate at which the separation shrinks is

-\frac{dr}{dt} = v - u\cos\alpha.

Integrating from the start, where $r=l$ , to the meeting instant $\tau$ , where $r=0$ , gives the total distance closed; the constant term $v$ integrates immediately, leaving the $\alpha$ -dependent piece untouched.

\int_0^\tau (v-u\cos\alpha)\,dt = v\tau - \int_0^\tau u\cos\alpha\,dt = l

v\tau - \displaystyle\int_0^\tau u\cos\alpha\,dt = l

(1)

Step 2The displacement of

A

along

B

’s line of motion

$A$ ’s velocity $\vect v$ points along $AB$ , which makes angle $\alpha$ with $\vect u$ ’s direction, so the component of $\vect v$ along $\vect u$ ’s direction is $v\cos\alpha$ . Because the initial line $AB$ is perpendicular to $\vect u$ (the given $\vect v_0\perp\vect u$ ), $A$ and $B$ start level with each other along $\vect u$ ’s direction, with the same coordinate there; since they also end together, at the same point, at time $\tau$ , their displacements along $\vect u$ ’s direction over $[0,\tau]$ must be equal too, and $B$ ’s is simply $u\tau$ (uniform motion in a straight line):

\int_0^\tau v\cos\alpha\,dt = u\tau.

\displaystyle\int_0^\tau v\cos\alpha\,dt = u\tau

(2)

Step 3Eliminate the integral of

\cos\alpha

The two boxed relations share the same integral $\int_0^\tau u\cos\alpha\,dt$ , up to the factor $u/v$ ; substituting (2) into (1) removes $\alpha$ entirely.

\begin{align*}\int_0^\tau u\cos\alpha\,dt &= \frac{u}{v}\int_0^\tau v\cos\alpha\,dt = \frac{u}{v}(u\tau) = \frac{u^2\tau}{v}\\ v\tau - \frac{u^2\tau}{v} &= l\\ \tau\left(\frac{v^2-u^2}{v}\right) &= l\end{align*}

\tau = \dfrac{vl}{v^2-u^2}

(3)

Check

Dimensionally, $(\text{m/s})(\text{m})/(\text{m}^2/\text{s}^2)=\text{s}$ , a time, as required. As $u\to0$ ( $B$ stationary), $\tau\to l/v$ , the ordinary time to cover a fixed distance $l$ at speed $v$ , exactly as it must when the target never moves. As $u\to v$ , $\tau\to\infty$ : a target as fast as the chaser, launched perpendicular to the chase, is never actually caught, consistent with $u<v$ being required for convergence.

Remark

Step 2’s trick, equating $A$ ’s net displacement along $\vect u$ to $B$ ’s net displacement along $\vect u$ , sidesteps ever solving for $\alpha(t)$ or for the pursuit curve’s actual shape. Pure pursuit curves are generally transcendental, yet the total time here falls out of two linear bookkeeping relations because both points must be at the very same location at $\tau$ , and a shared endpoint constrains the displacements without needing the path between them.