A particle P is moving with a constant speed u on a straight line that makes an angle theta with the

Q: Why can't we just use relative velocity in 1D?

Because the velocity of Q is constantly changing its direction to point towards P. This makes it a 2D pursuit problem where the line of sight is continuously rotating, so simple 1D relative velocity formulas don't apply directly.

Q: How did you know to take the dot product with P's velocity?

It's a classic mathematical trick for pursuit problems! By taking the dot product with $\vec{v}_P$, we naturally generate the term $v u \cos\alpha$, which perfectly matches the integral we get from the rate of change of distance equation.

Q: Why is the acceleration of Q purely centripetal in part (b)?

Acceleration has two components: tangential (changing speed) and centripetal (changing direction). Since Q moves with a constant speed $u$, its tangential acceleration is zero. Thus, all its acceleration goes into changing its direction to track P.

Q: What happens if $u > v$?

If P is faster than Q, Q will never catch P, and the distance between them will eventually increase to infinity. The equations we derived for catch time would yield a negative value, which is physically meaningless.

Q: Why does the minimum distance occur at $\alpha = 0$?

$\alpha$ is the angle between P's velocity and the line of sight. When $\alpha = 0$, P is moving directly along the line of sight away from Q. At this exact moment, their relative velocity along the line of sight is zero, marking the turning point from decreasing distance to increasing distance.

Question

A particle P is moving with a constant speed $u$ on a straight line that makes an angle $	heta$ with the positive $x$-direction of a coordinate system. When P crosses the $y$-axis at a point $(0, l)$, another particle Q starts from the origin and chases P with a uniform speed $v$ ($v > u$). The chaser Q always maintains its velocity vector towards the chased P.
(a) How long after Q starts from the origin, will it catch P?
(b) If both the chaser Q and the chased P move with equal speeds (i.e. $u = v$), what will be the minimum distance between them and what will be the maximum magnitude of acceleration of the chaser Q?

Sigma Percentile · Accepted Answer

### The Ultimate Pursuit: A Masterclass in Kinematics

Imagine a coordinate system where a particle P starts on the y-axis at a distance $l$ from the origin. It moves with a constant speed $u$ along a straight line making an angle $	heta$ with the positive x-axis. At the exact same moment, a chaser particle Q starts from the origin. Q is like a heat-seeking missile, always pointing its velocity vector directly at P, moving with a speed $v$. This is a classic pursuit problem, and solving it requires a blend of physical intuition and mathematical elegance.

### Analyzing the Rate of Approach

Let's look at the distance $s$ between them at any time $t$. How fast is this distance closing? Q is moving directly towards P with speed $v$, actively trying to reduce the distance. However, P is moving away at an angle. If $\alpha$ is the angle between P's velocity vector $\vec{v}_P$ and the line of sight $\vec{QP}$, P's velocity component along this line is $u \cos\alpha$.

Therefore, the net rate of change of distance is given by:

$$ \frac{ds}{dt} = u \cos\alpha - v $$

### The Genius Dot Product Trick

To find the time $T$ when Q catches P, we could try to set up complex differential equations for their trajectories. But there is a much more elegant way. If Q catches P at time $T$, their final position vectors must be exactly the same.

The position of P at time $T$ is its initial position plus its displacement: $\vec{r}_P(T) = l \hat{j} + \vec{v}_P T$. 
For Q, its position is the integral of its velocity vector over time: $\vec{r}_Q(T) = \int_0^T \vec{v}_Q dt$.

Equating them gives:

$$ \int_0^T \vec{v}_Q dt = l \hat{j} + \vec{v}_P T $$

Here comes the brilliant trick: we take the dot product of both sides with P's velocity vector $\vec{v}_P$. Why? Because the dot product of Q's velocity and $\vec{v}_P$ gives us $v u \cos\alpha$, which links perfectly with our distance equation!

$$ \int_0^T (\vec{v}_Q \cdot \vec{v}_P) dt = \vec{v}_P \cdot (l \hat{j}) + \vec{v}_P \cdot \vec{v}_P T $$

Evaluating the dot products on the right side, we get $u l \sin	heta + u^2 T$. So our equation becomes:

$$ v \int_0^T u \cos\alpha dt = u l \sin	heta + u^2 T $$

### Solving for the Catch Time

Now, let's bring back our distance equation. If we integrate $\frac{ds}{dt}$ from time $0$ to $T$, the distance goes from $l$ to $0$.

$$ \int_0^T (u \cos\alpha - v) dt = -l $$

Rearranging this, we find that $\int_0^T u \cos\alpha dt = v T - l$. We can now substitute this entire integral into our dot product equation:

$$ v (v T - l) = u l \sin	heta + u^2 T $$

Expanding and grouping the terms with $T$ on one side:

$$ T (v^2 - u^2) = l (v + u \sin	heta) $$

Dividing by the velocity difference, we get our final catch time:

$$ T = \frac{l (v + u \sin	heta)}{v^2 - u^2} $$

### The Tractrix Scenario: Equal Speeds

Now for part (b). What if their speeds are equal ($u = v$)? In this case, Q will never actually catch P. But how close does it get? Let's analyze the relative motion again. The rate of approach is now $-u(1 - \cos\alpha)$.

What about the rotation of the line of sight? The perpendicular velocity component of P is $u \sin\alpha$, which causes the line of sight of length $s$ to rotate at an angular rate. This gives us our second differential equation:

$$ s \frac{d\alpha}{dt} = -u \sin\alpha $$

Dividing the distance equation by the angle equation eliminates time completely:

$$ \frac{ds}{d\alpha} = s \frac{1 - \cos\alpha}{\sin\alpha} = s 	an(\frac{\alpha}{2}) $$

Integrating both sides yields $s = \frac{C}{\cos^2(\alpha/2)}$. Using the initial conditions at $t=0$ where $s=l$ and the line of sight is along the y-axis, we find the constant $C = l \frac{1 + \sin	heta}{2}$. The minimum distance occurs when the denominator is maximum (at $\alpha = 0$), which simply leaves us with the constant $C$!

### Maximum Acceleration of the Chaser

Finally, we need the maximum acceleration of Q. Since Q moves with a constant speed $u$, it has no tangential acceleration. Its acceleration is purely centripetal, given by $a_Q = u \frac{d\phi}{dt} = \frac{u^2 \sin\alpha}{s}$.

Substituting our formula for $s$, we get a function purely in terms of $\alpha$:

$$ a_Q = \frac{2 u^2}{C} \sin(\frac{\alpha}{2}) \cos^3(\frac{\alpha}{2}) $$

To find the maximum acceleration, we maximize the trigonometric part by taking its derivative and setting it to zero. This reveals the maximum occurs at $\alpha = \frac{\pi}{3}$ ($60^\circ$). Plugging this back in, we arrive at the final masterpiece:

$$ a_{max} = \frac{3\sqrt{3} v^2}{4 l (1 + \sin	heta)} $$

This problem is a phenomenal demonstration of how choosing the right mathematical tools—like dot products and relative velocity components—can turn a seemingly impossible calculus nightmare into an elegant symphony of logic.

Visualized Solution

Kinematic Setup

Rate of Approach

Position Vectors at Catch Time

The Dot Product Trick

Integrating the Distance Equation

Solving for Catch Time $T$

Equal Speeds: $u = v$

Differential Equation for Distance

Finding Minimum Distance

Acceleration of Chaser Q

Maximizing Acceleration

The Sigma Insight: Relative Velocity

The Ultimate Pursuit

Analyzing the Rate of Approach

The Genius Dot Product Trick

Solving for the Catch Time

The Tractrix Scenario

Maximum Acceleration of the Chaser

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