Two ships A and B can establish mutual communication when they are not more than 50 km apart. At mid

Q: Why do we use relative motion here? Can't we just find their individual positions and use the distance formula?

You absolutely can! If you write the coordinates of A as $(16t, 16t)$ and B as $(80, 4t)$, and apply the distance formula $\sqrt{(16t - 80)^2 + (16t - 4t)^2} \le 50$, you will get the exact same quadratic equation. Relative motion just provides a cleaner, more intuitive geometric visualization of one ship moving along a straight line while the other is stationary.

Q: How did we get the velocity components for Ship A?

Ship A is moving 'northeast', which means it makes an angle of $45^\circ$ with both the North (y-axis) and East (x-axis) directions. Using trigonometry, the x-component is $v \cos(45^\circ)$ and the y-component is $v \sin(45^\circ)$. Since $v = 16\sqrt{2}$ and $\cos(45^\circ) = \sin(45^\circ) = 1/\sqrt{2}$, both components become 16.

Q: Why did we square the distance inequality?

The distance formula involves a square root. Working with square roots in inequalities can be messy and prone to algebraic errors. Since distance is always positive, squaring both sides of the inequality $|\vec{r}| \le 50$ to get $|\vec{r}|^2 \le 2500$ is mathematically valid and eliminates the square root, making the expansion much easier.

Q: What does the shortest distance ($d_{min}$) represent physically?

The shortest distance is the perpendicular dropped from Ship A to the relative path of Ship B. If this distance was greater than 50 km, the ships would never come into communication range, and our quadratic equation would have imaginary roots!

Q: How do we convert 3.9 hours into hours and minutes?

The integer part, 3, represents the hours. The decimal part, 0.9, represents a fraction of an hour. Since there are 60 minutes in an hour, you multiply $0.9 \times 60$ to get 54 minutes. So, 3.9 hours is exactly 3 hours and 54 minutes.

Question

Two ships A and B can establish mutual communication when they are not more than 50 km apart. At midnight, the ship B moving towards the north with a velocity 4 km/h passes a location 80 km east of ship A that is moving with velocity $16\sqrt{2}$ km/h towards the northeast. Find the time interval during which they were in communication.

Sigma Percentile · Accepted Answer

### The Midnight Setup
Imagine the vast, dark ocean at exactly midnight. Two ships, A and B, are navigating the waters. Ship A is our reference point, so let's anchor it at the origin of our coordinate system, $(0,0)$. Ship B is currently 80 km due east, sitting at the coordinates $(80,0)$.

The crucial constraint here is their communication equipment. They can only talk to each other if the distance between them is 50 km or less. Geometrically, this means Ship B must enter a circle of radius 50 km centered around Ship A.

### The Velocity Vectors
Both ships are on the move, which makes tracking their distance a dynamic challenge. 
Ship A is slicing through the water towards the northeast at a speed of $16\sqrt{2} 	ext{ km/h}$. Because "northeast" implies a $45^\circ$ angle, we can break this velocity into its $x$ and $y$ components. Using basic trigonometry, both components are $16\sqrt{2} \cos(45^\circ) = 16 	ext{ km/h}$. Thus, $\vec{v}_A = 16\hat{i} + 16\hat{j}$.

Ship B has a simpler trajectory. It is heading straight north at $4 	ext{ km/h}$, giving it a velocity vector of $\vec{v}_B = 4\hat{j}$.

### The Power of Relative Motion
Tracking two moving objects simultaneously is a recipe for algebraic headaches. Instead, we use a brilliant physicist's trick: **Relative Motion**. 
By subtracting Ship A's velocity from Ship B's, we can pretend Ship A is completely stationary and observe how Ship B moves relative to it.

Let's calculate this relative velocity:
$$ \vec{v}_{BA} = \vec{v}_B - \vec{v}_A $$
$$ \vec{v}_{BA} = (0\hat{i} + 4\hat{j}) - (16\hat{i} + 16\hat{j}) = -16\hat{i} - 12\hat{j} $$

This tells us a beautiful story: from the perspective of Ship A, Ship B is moving southwest along a perfectly straight line!

### The Master Equation
Now we need to find the position of Ship B relative to Ship A at any given time $t$. We use the standard kinematic equation for constant velocity:
$$ \vec{r}_{BA}(t) = \vec{r}_{B0} + \vec{v}_{BA}t $$
Substituting our known values:
$$ \vec{r}_{BA}(t) = 80\hat{i} + (-16\hat{i} - 12\hat{j})t $$
$$ \vec{r}_{BA}(t) = (80 - 16t)\hat{i} - 12t\hat{j} $$

### The Communication Constraint
For the ships to communicate, the magnitude of this relative position vector must be less than or equal to 50 km. 
$$ |\vec{r}_{BA}(t)| \le 50 $$

To avoid dealing with messy square roots, we square both sides of the inequality:
$$ |\vec{r}_{BA}(t)|^2 \le 2500 $$

Now, we plug in the $x$ and $y$ components of our position vector:
$$ (80 - 16t)^2 + (-12t)^2 \le 2500 $$

### The Final Calculation
Let's expand this carefully. 
$$ 6400 - 2560t + 256t^2 + 144t^2 \le 2500 $$
Combining the $t^2$ terms gives us $400t^2$. Bringing the 2500 over to the left side yields a clean quadratic inequality:
$$ 400t^2 - 2560t + 3900 \le 0 $$

To make the math friendlier, let's divide the entire equation by 20:
$$ 20t^2 - 128t + 195 \le 0 $$

Using the quadratic formula, $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find the roots:
$$ t = \frac{128 \pm \sqrt{16384 - 15600}}{40} = \frac{128 \pm 28}{40} $$

This gives us two critical moments in time:
$t_1 = 2.5 	ext{ hours}$ (when Ship B enters the communication circle)
$t_2 = 3.9 	ext{ hours}$ (when Ship B exits the communication circle)

Since the clock started at midnight, $2.5 	ext{ hours}$ is exactly **2:30 A.M.** 
For the exit time, $3.9 	ext{ hours}$ is 3 hours and $0.9 	imes 60 = 54 	ext{ minutes}$, which is **3:54 A.M.**

Thus, the ships can successfully communicate during the time interval from **2:30 A.M. to 3:54 A.M.**

Visualized Solution

$I ni t ia l S e t u p & C oor d ina t e S y s t e m$

$V e l oc i t y V ec t or s$

$C o n ce pt o f R e l a t i v e M o t i o n$

$P os i t i o n V ec t or o f B r e l a t i v e t o A$

$A ppl y in g t h e C o mm u ni c a t i o n C o n s t r ain t$

$E x p an d in g t h e E q u a t i o n$

$S o l v in g t h e Q u a d r a t i c I n e q u a l i t y$

$F ina l T im e I n t er v a l$

The Sigma Insight: Relative Velocity

The Midnight Setup Imagine the vast, dark ocean at exactly midnight

The Velocity Vectors Both ships are on the move, which makes tracking their distance a dynamic challenge

The Power of Relative Motion Tracking two moving objects simultaneously is a recipe for algebraic headaches

The Master Equation Now we need to find the position of Ship B relative to Ship A at any given time $t$

The Communication Constraint

The Final Calculation Let's expand this carefully.
$6400 - 2560 t + 256 t^{2} + 144 t^{2} \leq 2500$
Combining the $t^{2}$ terms gives us $400 t^{2}$

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup & Coordinate System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity Vectors

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Concept of Relative Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position Vector of B relative to A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Communication Constraint

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expanding the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving the Quadratic Inequality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Time Interval