Physics Solution: Pursuit Problem (Projection Method)
1. Definitions and Setup
Let the velocity of the chaser Q be $\vec{v}$ (magnitude $v$) directed towards P. Let the velocity of the target P be $\vec{u}$ (magnitude $u$) at a constant angle $\theta$ to the horizontal.
Let $\phi$ be the angle between the velocity vector of Q ($\vec{v}$) and the velocity vector of P ($\vec{u}$).
Let the velocity of the chaser Q be $\vec{v}$ (magnitude $v$) directed towards P. Let the velocity of the target P be $\vec{u}$ (magnitude $u$) at a constant angle $\theta$ to the horizontal.
Let $\phi$ be the angle between the velocity vector of Q ($\vec{v}$) and the velocity vector of P ($\vec{u}$).
2. Rate of Approach Equation
The rate at which the distance $r$ between Q and P decreases is given by the relative velocity along the line joining them: $$ -\frac{dr}{dt} = v – u \cos\phi $$ Integrating from $t=0$ to $t=\tau$: $$ \int_{l}^{0} -dr = \int_{0}^{\tau} (v – u \cos\phi) dt $$ $$ l = v\tau – u \int_{0}^{\tau} \cos\phi \, dt \quad \dots(1) $$
The rate at which the distance $r$ between Q and P decreases is given by the relative velocity along the line joining them: $$ -\frac{dr}{dt} = v – u \cos\phi $$ Integrating from $t=0$ to $t=\tau$: $$ \int_{l}^{0} -dr = \int_{0}^{\tau} (v – u \cos\phi) dt $$ $$ l = v\tau – u \int_{0}^{\tau} \cos\phi \, dt \quad \dots(1) $$
3. Displacement Projection Equation
We project the displacements of both particles along the direction of P’s constant velocity vector $\vec{u}$.
The displacement of Q projected along $\vec{u}$ must equal the displacement of P along $\vec{u}$ plus the projection of the initial separation vector.
* Displacement of P along $\vec{u}$: $u\tau$ * Projection of initial position difference along $\vec{u}$: The initial separation is $l$ along the y-axis. The angle between the y-axis and $\vec{u}$ is $(90^\circ – \theta)$. Thus, the projection is $l \cos(90^\circ – \theta) = l \sin\theta$. * Displacement of Q along $\vec{u}$: $\int_{0}^{\tau} v \cos\phi \, dt$ Equating these: $$ \int_{0}^{\tau} v \cos\phi \, dt = u\tau + l \sin\theta \quad \dots(2) $$
We project the displacements of both particles along the direction of P’s constant velocity vector $\vec{u}$.
The displacement of Q projected along $\vec{u}$ must equal the displacement of P along $\vec{u}$ plus the projection of the initial separation vector.
* Displacement of P along $\vec{u}$: $u\tau$ * Projection of initial position difference along $\vec{u}$: The initial separation is $l$ along the y-axis. The angle between the y-axis and $\vec{u}$ is $(90^\circ – \theta)$. Thus, the projection is $l \cos(90^\circ – \theta) = l \sin\theta$. * Displacement of Q along $\vec{u}$: $\int_{0}^{\tau} v \cos\phi \, dt$ Equating these: $$ \int_{0}^{\tau} v \cos\phi \, dt = u\tau + l \sin\theta \quad \dots(2) $$
4. Solving for Time ($\tau$)
From equation (1), we can express the integral term: $$ \int_{0}^{\tau} \cos\phi \, dt = \frac{v\tau – l}{u} $$ Substitute this into equation (2): $$ v \left( \frac{v\tau – l}{u} \right) = u\tau + l \sin\theta $$ Multiply by $u$: $$ v(v\tau – l) = u(u\tau + l \sin\theta) $$ $$ v^2\tau – vl = u^2\tau + ul \sin\theta $$ $$ (v^2 – u^2)\tau = l(v + u \sin\theta) $$
From equation (1), we can express the integral term: $$ \int_{0}^{\tau} \cos\phi \, dt = \frac{v\tau – l}{u} $$ Substitute this into equation (2): $$ v \left( \frac{v\tau – l}{u} \right) = u\tau + l \sin\theta $$ Multiply by $u$: $$ v(v\tau – l) = u(u\tau + l \sin\theta) $$ $$ v^2\tau – vl = u^2\tau + ul \sin\theta $$ $$ (v^2 – u^2)\tau = l(v + u \sin\theta) $$
$$ \tau = \frac{l(v + u \sin\theta)}{v^2 – u^2} $$
(b) Minimum Distance when u = v
If $u=v$, the particle Q never catches P. We need to find the minimum distance $r_{min}$.At minimum distance, the velocity of approach is zero: $$ \frac{dr}{dt} = 0 \implies v – u \cos\phi = 0 $$ Since $u=v$, this implies $\cos\phi = 1 \implies \phi = 0$. (The velocity vectors become parallel).
We return to the general forms of equations (1) and (2) for any time $t$ and distance $r$. From Approach (Eq 1): $$ r = l – vt + u \int \cos\phi \, dt \implies \int \cos\phi \, dt = \frac{r – l + vt}{u} $$ From Projection (Eq 2, modified for non-zero $r$): The projection equation relates positions. The projection of current separation $r$ along $\vec{u}$ is $r \cos\phi$. $$ \int v \cos\phi \, dt = u t + l \sin\theta – r \cos\phi $$ Substituting the integral expression into the projection equation: $$ v \left( \frac{r – l + vt}{u} \right) = ut + l \sin\theta – r \cos\phi $$ Set $u=v$ and $\cos\phi = 1$: $$ v \left( \frac{r – l + vt}{v} \right) = vt + l \sin\theta – r(1) $$ $$ r – l + vt = vt + l \sin\theta – r $$ $$ 2r = l + l \sin\theta $$
Minimum Distance ($r_{min}$):
$$ r_{min} = \frac{l(1 + \sin\theta)}{2} $$