Minimum Separation in 2D Motion

1. The Fundamental Condition for Minimum Separation

Imagine two ships, A and B, moving on the sea. The distance between them, $S$, changes with time. Initially, they might be moving towards each other, so $S$ decreases. At some point, they reach a closest approach, and then the distance begins to increase. The moment of minimum separation is the turning point where the distance stops decreasing and starts increasing.

From calculus, you know that a function is at a minimum when its first derivative is zero. So, the distance $S$ is minimum when $\frac{dS}{dt}=0$.

The quantity $\frac{dS}{dt}$ represents the rate of change of separation. Its negative, $-\frac{dS}{dt}$, is what we call the velocity of approach.

• If $\frac{dS}{dt}<0$, the objects are approaching each other.
• If $\frac{dS}{dt}>0$, the objects are separating from each other.
• If $\frac{dS}{dt}=0$, the distance between them is momentarily constant, which corresponds to a local minimum (or maximum).

Therefore, the condition for minimum separation is that the velocity of approach is zero.

Let's formalize this using vectors, which provides the most elegant and general proof.
Let ${\vec{r}}_A(t)$ and ${\vec{r}}_B(t)$ be the position vectors of the two objects. The relative position vector is ${\vec{r}}_{B/A}(t)={\vec{r}}_B(t)-{\vec{r}}_A(t)$. The squared distance between them is:
$S^2(t)=|{\vec{r}}_{B/A}(t){|}^2={\vec{r}}_{B/A}(t)\cdot {\vec{r}}_{B/A}(t)$
To find the minimum distance, we differentiate $S^2$ with respect to time and set the derivative to zero. (Minimizing $S$ is equivalent to minimizing $S^2$, which avoids dealing with square roots).

Using the product rule for dot products:
$\frac{d(S^2)}{dt}=\frac{d}{dt}({\vec{r}}_{B/A}\cdot {\vec{r}}_{B/A})=\frac{d{\vec{r}}_{B/A}}{dt}\cdot {\vec{r}}_{B/A}+{\vec{r}}_{B/A}\cdot \frac{d{\vec{r}}_{B/A}}{dt}=2{\vec{r}}_{B/A}\cdot \frac{d{\vec{r}}_{B/A}}{dt}$
Since $\frac{d{\vec{r}}_{B/A}}{dt}={\vec{v}}_{B/A}$ (the relative velocity), we get:
$\frac{d(S^2)}{dt}=2{\vec{r}}_{B/A}\cdot {\vec{v}}_{B/A}$
Setting this to zero for minimum separation gives our fundamental condition:
${\vec{r}}_{B/A}\cdot {\vec{v}}_{B/A}=0$
This equation states that at the instant of closest approach, the relative position vector is perpendicular to the relative velocity vector. This is a profound and useful result.

2. Method 1: The Relative Frame (Geometric) Approach

For objects in uniform motion ($\vec{a}=0$), the most intuitive method is often to switch into the reference frame of one of the objects. Since ${\vec{a}}_{B/A}={\vec{a}}_B-{\vec{a}}_A=\vec{0}-\vec{0}=\vec{0}$, the relative motion is also uniform.

The Strategy:

1. Fix one object: Consider the frame of reference of object A. In this frame, A is stationary at the origin.
2. Find relative velocity: Calculate the velocity of object B relative to A: ${\vec{v}}_{B/A}={\vec{v}}_B-{\vec{v}}_A$.
3. Find initial relative position: Determine the initial position of B relative to A: ${\vec{r}}_{B/A}(0)$.
4. Analyze the relative motion: In A's frame, B starts at ${\vec{r}}_{B/A}(0)$ and moves in a straight line with constant velocity ${\vec{v}}_{B/A}$.
5. Find the minimum distance: The shortest distance from the stationary object A to the path of B is the perpendicular distance from the origin to the line of motion.

This process transforms a two-body problem into a simple geometry problem of finding the distance from a point to a line.

3. Method 2: The Calculus (Analytical) Approach

While the geometric method is elegant, a direct application of calculus is a robust and powerful alternative that leverages your mathematical strengths. This was "Method II" in Example 28 from the PDF you read.

The Strategy:

1. Establish a coordinate system: Define an (x, y) coordinate system.
2. Write positions vs. time: For each object, write its coordinates as a function of time. For uniform motion, these will be linear functions:
   • Object A: $x_A(t)=x_{A0}+v_{Ax}t$, $y_A(t)=y_{A0}+v_{Ay}t$
   • Object B: $x_B(t)=x_{B0}+v_{Bx}t$, $y_B(t)=y_{B0}+v_{By}t$
3. Form the distance function: Write the expression for the square of the distance between them:
   $S^2(t)=(x_B(t)-x_A(t){)}^2+(y_B(t)-y_A(t){)}^2$
   Substituting the linear functions for the coordinates will result in $S^2(t)$ being a quadratic function of time, i.e., $S^2(t)=a{t}^2+bt+c$.
4. Minimize the function: Find the time $t_{min}$ at which $S^2(t)$ is minimum. You can do this by:
   • Setting the derivative to zero: $\frac{d(S^2)}{dt}=2at+b=0⟹t=-b/(2a)$.
   • Or, for a parabola, the vertex is at $t=-b/(2a)$.
5. Calculate minimum distance: Substitute $t_{min}$ back into the expression for $S(t)$ to find $S_{min}$.

This method is algorithmic and guaranteed to work, though it can sometimes involve more algebra than the geometric approach.

4. General Vector Formulas and Their Derivation

For a deeper understanding, let's derive the general vector formulas for the time of closest approach and the minimum distance. These are mentioned in the RELATIVE MOTION PDF resource.

Let ${\vec{r}}_{12}={\vec{r}}_2(0)-{\vec{r}}_1(0)$ be the initial relative position vector and ${\vec{v}}_{12}={\vec{v}}_2-{\vec{v}}_1$ be the constant relative velocity.

Derivation of Time of Closest Approach ($t_{min}$)
Our fundamental condition is that at time $t_{min}$, the relative position vector ${\vec{r}}_{12}(t_{min})$ is perpendicular to the relative velocity ${\vec{v}}_{12}$.
${\vec{r}}_{12}(t_{min})\cdot {\vec{v}}_{12}=0$
The relative position at any time $t$ is ${\vec{r}}_{12}(t)={\vec{r}}_{12}(0)+{\vec{v}}_{12}t$. Substituting this into our condition:
$({\vec{r}}_{12}(0)+{\vec{v}}_{12}{t}_{min})\cdot {\vec{v}}_{12}=0$
${\vec{r}}_{12}(0)\cdot {\vec{v}}_{12}+t_{min}({\vec{v}}_{12}\cdot {\vec{v}}_{12})=0$
${\vec{r}}_{12}\cdot {\vec{v}}_{12}+t_{min}|{\vec{v}}_{12}{|}^2=0$
Solving for $t_{min}$ gives:
$t_{min}=-\frac{{\vec{r}}_{12}\cdot {\vec{v}}_{12}}{|{\vec{v}}_{12}{|}^2}$

Derivation of Minimum Distance ($d_{min}$)
The minimum distance is the magnitude of the relative position vector at $t_{min}$, i.e., $d_{min}=|{\vec{r}}_{12}(t_{min})|$. A more elegant derivation uses the cross product.

Consider the parallelogram formed by the vectors ${\vec{r}}_{12}(0)$ and ${\vec{v}}_{12}$.
The magnitude of their cross product gives the area of this parallelogram: $Area=|{\vec{r}}_{12}\times {\vec{v}}_{12}|$.
From a geometric perspective, this area can also be calculated as base times height. In the relative frame, the "base" is the path length along the direction of relative velocity, and the "height" is the perpendicular distance of closest approach, $d_{min}$. The length of the base vector is $|{\vec{v}}_{12}|$.
So, $Area=|{\vec{v}}_{12}|\cdot d_{min}$.
Equating the two expressions for the area:
$|{\vec{v}}_{12}|\cdot d_{min}=|{\vec{r}}_{12}\times {\vec{v}}_{12}|$
This gives the beautiful formula for the shortest distance:
$d_{min}=\frac{|{\vec{r}}_{12}\times {\vec{v}}_{12}|}{|{\vec{v}}_{12}|}$

5. Advanced Application: Minimum Separation Between Projectiles

What happens if the objects are accelerating, like two projectiles fired simultaneously? The calculus method becomes more complex. Can we still use the relative motion approach?

Let projectile A and projectile B have velocities ${\vec{v}}_A$ and ${\vec{v}}_B$ and accelerations ${\vec{a}}_A$ and ${\vec{a}}_B$.
The relative acceleration is ${\vec{a}}_{B/A}={\vec{a}}_B-{\vec{a}}_A$.
For two projectiles moving under gravity (neglecting air resistance), ${\vec{a}}_A={\vec{a}}_B=\vec{g}=(0,-g)$.
Therefore, the relative acceleration is:
${\vec{a}}_{B/A}=\vec{g}-\vec{g}=\vec{0}$
This is a crucial insight! From the perspective of one projectile, the other projectile moves with constant velocity. Gravity's effect is cancelled out in the relative frame. This means we can apply the exact same geometric method we developed for uniform motion.

Test your understanding!

Consider the problem shown in the image below. Using two different methods (e.g., the geometric/relative frame method and the calculus method), calculate the shortest distance between particles A and B.

Show answer

Let's solve this using both methods.

Method 1: Relative Frame (Geometric)

1. Reference Frame: Let's analyze the motion of B with respect to A. A becomes stationary at its initial position $(-30,0)$. Wait, it's easier to place A at the origin of the relative frame.
2. Initial Relative Position: ${\vec{r}}_{B/A}(0)={\vec{r}}_B(0)-{\vec{r}}_A(0)=(0\hat{i}-40\hat{j})-(-30\hat{i}+0\hat{j})=30\hat{i}-40\hat{j}$ m.
3. Relative Velocity: ${\vec{v}}_{B/A}={\vec{v}}_B-{\vec{v}}_A=(20\hat{j})-(10\hat{i})=-10\hat{i}+20\hat{j}$ m/s.
4. Geometric Setup: In A's frame, B starts at $(30,-40)$ and moves with a constant velocity $(-10,20)$. We need to find the perpendicular distance from the origin (A's position) to the line of B's motion.
5. Time of Closest Approach: We use the formula $t_{min}=-\frac{{\vec{r}}_{B/A}\cdot {\vec{v}}_{B/A}}{|{\vec{v}}_{B/A}{|}^2}$.
   • ${\vec{r}}_{B/A}\cdot {\vec{v}}_{B/A}=(30)(-10)+(-40)(20)=-300-800=-1100$
   • $|{\vec{v}}_{B/A}{|}^2=(-10{)}^2+(20{)}^2=100+400=500$
   • $t_{min}=-\frac{-1100}{500}=\frac{11}{5}=2.2$ s.
6. Minimum Distance: Now find the relative position at $t=2.2$ s.
   • ${\vec{r}}_{B/A}(2.2)={\vec{r}}_{B/A}(0)+{\vec{v}}_{B/A}t=(30\hat{i}-40\hat{j})+(-10\hat{i}+20\hat{j})(2.2)$
   • ${\vec{r}}_{B/A}(2.2)=(30-22)\hat{i}+(-40+44)\hat{j}=8\hat{i}+4\hat{j}$
   • $d_{min}=|{\vec{r}}_{B/A}(2.2)|=\sqrt{8^2+4^2}=\sqrt{64+16}=\sqrt{80}=4\sqrt{5}$ m.

Method 2: Calculus

1. Positions vs. Time:
   • A: $x_A(t)=-30+10t$, $y_A(t)=0$
   • B: $x_B(t)=0$, $y_B(t)=-40+20t$
2. Squared Distance Function:
   • $S^2(t)=(x_B-x_A{)}^2+(y_B-y_A{)}^2$
   • $S^2(t)=(0-(-30+10t){)}^2+(-40+20t-0{)}^2$
   • $S^2(t)=(30-10t{)}^2+(20t-40{)}^2$
   • $S^2(t)=(900-600t+100t^2)+(400t^2-1600t+1600)$
   • $S^2(t)=500t^2-2200t+2500$
3. Minimize: Set the derivative to zero.
   • $\frac{d(S^2)}{dt}=1000t-2200=0$
   • $t=\frac{2200}{1000}=2.2$ s. (This matches Method 1).
4. Calculate Minimum Distance: Substitute $t=2.2$ s back into $S^2(t)$.
   • $S^2(2.2)=500(2.2{)}^2-2200(2.2)+2500$
   • $S^2(2.2)=500(4.84)-4840+2500=2420-4840+2500=80$
   • $S_{min}=\sqrt{80}=4\sqrt{5}$ m.

Both methods give the same result, confirming our understanding.

Conclusion

Today we have developed a comprehensive framework for one of the most important problems in 2D kinematics. By starting from the first principle of minimizing distance ($\frac{dS}{dt}=0$), we derived the powerful and elegant condition that the relative position vector must be perpendicular to the relative velocity vector at the point of closest approach.

Key Takeaways:

• The fundamental condition for minimum separation between two objects is ${\vec{r}}_{relative}\cdot {\vec{v}}_{relative}=0$.
• For objects in uniform motion, this problem can be solved geometrically by switching to a relative frame where one object is stationary and the other moves in a straight line. The minimum distance is the perpendicular from the stationary object to this line.
• A robust alternative is the analytical (calculus) method, where you write the squared distance as a quadratic function of time, $S^2(t)$, and find its minimum.
• The insight that relative acceleration between two projectiles under gravity is zero allows us to apply the same uniform motion techniques to this more complex scenario.

Preview of the Next Lesson:
The condition for minimum separation naturally leads to a special case: what if the minimum separation is zero? This is the condition for a collision or interception. In our next lesson, we will derive and apply the condition for collision or interception between two objects using relative velocity vectors. We will find that for a collision to occur, the relative velocity vector must point directly along the initial relative position vector.