Question

Point $A$ moves uniformly with velocity $v=13\text{m/s}$ so that the vector $u=5\text{m/s}$ is continually aimed at point $B$ which in 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 $l=12\text{m}$. How soon will the points converge?

• A. $\frac{12}{13}\text{sec}$
• B. $\frac{5}{12}\text{sec}$
• C. $\frac{13}{12}\text{sec}$
• D. They will not Converge

Answer 1

The correct option is C $\frac{13}{12}\text{sec}$


At $t=0$, points are at $A$ & $B$
At some time ${}^{\prime }{t}^{\prime }$, position of points are $A^{\prime }$ & $B^{\prime }$ respectively.
Let $B{B}^{\prime }=$ distance travelled by $B$ in time $t_0=u{t}_0...\left(1\right)$
For point $A$, ${\int }_0^{t_0}v\cos \alpha dt=B{B}^{\prime }=u{t}_0$
$\Rightarrow {\int }_0^{t_0}\cos \alpha dt=\frac{u{t}_0}{v}...\left(2\right)$
${}^{\prime }{t}_0^{\prime }$ is the time after which they will converge.
Relative velocity of approach of $A$ to $B$ in time ${}^{\prime }{t}^{\prime }$
$v_0=v-u\cos \alpha$
For converging the points $A$ & $B$,
$l={\int }_0^{t_0}{v}_{0}dt={\int }_0^{t_0}(v-u\cos \alpha )dt$
$={\int }_0^{t_0}vdt-{\int }_0^{t_0}u\cos \alpha dt=v{t}_0-\frac{u^2{t}_0}{v}$.....from $\left(2\right)$
$\therefore t_0=\frac{lv}{v^2-u^2}=\frac{12\times 13}{{13}^2-5^2}=\frac{13}{12}\text{sec}$.
Hence, option $\left(C\right)$ is correct.