Question

Point $A$ moves uniformly with velocity $v$ so that the vector $v$ is continually "aimed" at point $B$ which in its turn moves rectilinearly and uniformly with velocity $u<v$. At the initial moment of time $v\perp u$ and the points are separated by a distance $l$. The time at which the points converge is given as $T=\frac{xvl}{v^2-u^2}$. Find $x$.

Answer 1

Let us locate the points $A$ and $B$ at an arbitrary instant of time (as shown in the figure).
If $A$ and $B$ are separated by the distance $s$ at this moment, then the points converge or point $A$ approaches $B$ with velocity $\frac{-ds}{dt}=v-u\cos \alpha$ where angle $\alpha$ varies with time.
On integrating,
$-{\int }_l^{0}ds={\int }_0^T(v-u\cos \alpha )dt$,
(where $T$ is the sought time.)
or $l={\int }_0^T(v-u\cos \alpha )dt$.............. (1)
As both $A$ and $B$ cover the same distance in $x-$ direction during the sought time interval, so the other condition which is required, can be obtained by the equation
$\Delta x=\int v_{x}dt$
So, $uT={\int }_0^{T}v\cos \alpha dt$................ (2)
Solving (1) and (2), we get $T=\frac{vl}{v^2-u^2}$
One can see that if $u=v$, or $u<v$, point $A$ cannot catch $B$.