Two Particles Meeting Under Acceleration
Let the initial separation be $d$.
Particle 1: Velocity $u_1$ (right), Acceleration $a_1$ (left).
Particle 2: Velocity $u_2$ (left), Acceleration $a_2$ (right).
Relative Motion Analysis
We analyze the motion of Particle 1 relative to Particle 2.
Relative Initial Velocity: $u_{rel} = u_1 – (-u_2) = u_1 + u_2$ (Closing in).
Relative Acceleration: $a_{rel} = (-a_1) – a_2 = -(a_1 + a_2)$ (Separating/Retarding).
The relative displacement $S_{rel}$ must be $d$ (initial separation) for them to meet. However, since we treat P2 as stationary at origin, P1 starts at $x=-d$ and must reach $0$.
Better convention: Let relative position $x(t)$ start at $0$. They meet when $x(t) = d$.
Equation of motion:
$$ d = u_{rel} t – \frac{1}{2} |a_{rel}| t^2 $$
$$ d = (u_1 + u_2)t – \frac{1}{2}(a_1 + a_2)t^2 $$
Rearranging into a quadratic equation for $t$:
$$ \frac{1}{2}(a_1 + a_2)t^2 – (u_1 + u_2)t + d = 0 $$
Solving for Separation
Since they meet twice, there are two distinct roots $t_1$ and $t_2$. The difference between roots is given as $\Delta t$.
For a quadratic $At^2 + Bt + C = 0$, the difference of roots is $\frac{\sqrt{B^2 – 4AC}}{A}$.
Here: $A = \frac{1}{2}(a_1 + a_2)$, $B = -(u_1 + u_2)$, $C = d$.
Squaring both sides:
$$ (\Delta t)^2 \frac{(a_1+a_2)^2}{4} = (u_1+u_2)^2 – 2(a_1+a_2)d $$ $$ 2(a_1+a_2)d = (u_1+u_2)^2 – \frac{(a_1+a_2)^2 (\Delta t)^2}{4} $$ $$ d = \frac{(u_1+u_2)^2}{2(a_1+a_2)} – \frac{(a_1+a_2)(\Delta t)^2}{8} $$Answer: $\frac{(u_1+u_2)^2}{2(a_1+a_2)} – \frac{(a_1+a_2)(\Delta t)^2}{8}$