Question

A situation is shown in which two objects $A$ and $B$ start their motion from same point in same direction. The graph of their velocities against time is drawn. $u_A$ and $u_B$ are the initial velocities of $A$ and $B$ respectively. $T$ is the time at which their velocities become equal after start of motion.

If the value of $T$ is $4\text{s}$, then the time after which $A$ will meet $B$ is

• A. $12\text{s}$
• B. $6\text{s}$
• C. $8\text{s}$
• D. data insufficient

Answer 1

The correct option is C $8\text{s}$

Let the particles meet at time $t$
i.e., displacement of the particles are equal and which is possible when

Area of $\Delta ACB=$ Area of $\Delta A^{\prime }{B}^{\prime }C$
[Area $OBC{A}^{\prime }RO$ being common]
or Area of $\Delta APC=$ Area $A^{\prime }{P}^{\prime }C$
$\frac{1}{2}AP\times PC=\frac{1}{2}{A}^{\prime }{P}^{\prime }\times P^{\prime }C$
or $AP\times PC=A^{\prime }{P}^{\prime }\times P^{\prime }C$
or $\left(PC\tan \theta \right)PC=\left(P^{\prime }C\tan \theta \right)P^{\prime }C$
or $PC=P^{\prime }{C}^{\prime }$
$\therefore OR=PC+P^{\prime }{C}^{\prime }=2PC=8\text{s}$
Option (c) is correct.