Question

Two particles are originally placed at $P$ and $Q$ distant $d$ apart. At $t=0$ they start moving such that velocity $\overrightarrow{v}$ of $P$ is aimed towards $Q$ and velocity $\overrightarrow{u}$ of $Q$ is perpendicular to $\overrightarrow{v}$. The two projectiles meet at time $T=$

• A. $\frac{(v^2-u^2)d^2}{v^{3}d}$
• B. $\frac{(v+u)d}{v^2}$
• C. $\frac{v(v-u)}{d}$
• D. $\frac{vd}{(v^2-u^2)}$

Answer 1

Answer: (D)

$\frac{vd}{(v^2-u^2)}$

Relative velocity of $P$ and $Q$ is $(v-u\cos \theta )$.

The particles will meet when:
${\int }_0^{\pi }(v-u\cos \theta )dt=d$ and ${\int }_0^{\pi }v\cos \theta dt=d$
and ${\int }_0^{\pi }\cos \theta dt=\frac{uT}{v}$

$\Rightarrow vT-u\frac{uT}{v}=d$

$\Rightarrow (v^2-u^2)T=vd$
$\Rightarrow T=\frac{vd}{v^2-u^2}$