Ship A is sailing towards north-east with velocity v = 30i + 50j km/h, where i points east and j nor

Q: Why do we use relative motion here instead of just writing equations for both ships?

Writing equations for both ships requires you to find the distance between two moving points, which leads to a messy quadratic equation in time. By using relative motion, we freeze one ship at the origin. The problem then beautifully simplifies to finding the perpendicular distance from the origin to a single straight line.

Q: How is the formula $t = -\frac{\mathbf{r}_{BA} \cdot \mathbf{v}_{BA}}{|\mathbf{v}_{BA}|^2}$ derived?

Let the relative position at time $t$ be $\mathbf{r}(t) = \mathbf{r}_0 + \mathbf{v}t$. The distance squared is $D^2 = |\mathbf{r}(t)|^2 = \mathbf{r}_0 \cdot \mathbf{r}_0 + 2(\mathbf{r}_0 \cdot \mathbf{v})t + |\mathbf{v}|^2 t^2$. To find the minimum distance, we take the derivative of $D^2$ with respect to $t$ and set it to zero: $\frac{d(D^2)}{dt} = 2(\mathbf{r}_0 \cdot \mathbf{v}) + 2|\mathbf{v}|^2 t = 0$. Solving for $t$ gives the formula.

Q: What does the negative sign in the time formula mean?

The dot product $\mathbf{r}_{BA} \cdot \mathbf{v}_{BA}$ is negative because the angle between the initial relative position vector and the relative velocity vector is obtuse (greater than 90 degrees). This means the ship is moving *towards* the point of closest approach. The negative sign in the formula cancels this out to give a positive time.

Q: Can we solve this using calculus instead of vectors?

Absolutely! You can write the coordinates of Ship A as $(30t, 50t)$ and Ship B as $(80 - 10t, 150)$. The distance squared is $D^2 = (80 - 40t)^2 + (150 - 50t)^2$. Differentiating this with respect to $t$ and equating to zero will give you the exact same result, $t = 107/41$.

Q: How would we find the actual minimum distance?

Once you have the time $t = 2.6$ h, you can substitute it back into the relative position equation $\mathbf{r}(t) = \mathbf{r}_{BA} + \mathbf{v}_{BA}t$ and find its magnitude. Alternatively, you can use the cross product formula $d_{\text{min}} = \frac{|\mathbf{r}_{BA} \times \mathbf{v}_{BA}|}{|\mathbf{v}_{BA}|}$, which directly gives the perpendicular distance.

Question

Ship A is sailing towards north-east with velocity $\mathbf{v} = 30\hat{\mathbf{i}} + 50\hat{\mathbf{j}}$ km/h, where $\hat{\mathbf{i}}$ points east and $\hat{\mathbf{j}}$ north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/h. A will be at minimum distance from B in

Options:

(A) 4.2 h

(B) 2.6 h

(C) 3.2 h

(D) 2.2 h

Sigma Percentile · Accepted Answer

### The Illusion of Two Moving Bodies
Imagine standing on the deck of a ship, looking out at the vast, open ocean. You see another ship in the distance, and both of you are moving. Trying to figure out exactly when you will be closest to each other by tracking both paths simultaneously is a geometric nightmare.

But what if you could freeze your ship in place? This is the magic of **relative motion**. By shifting our perspective to Ship A's frame of reference, we transform a complex two-body problem into a simple one-body problem. Ship A becomes the stationary origin, and Ship B does all the moving.

### Setting the Stage
Let's define our initial conditions. Ship A is at the origin, so its position vector is $\mathbf{r}_A = 0\hat{\mathbf{i}} + 0\hat{\mathbf{j}}$. It is sailing North-East with a velocity $\mathbf{v}_A = 30\hat{\mathbf{i}} + 50\hat{\mathbf{j}}$ km/h.

Ship B is located 80 km East and 150 km North, giving it an initial position vector $\mathbf{r}_B = 80\hat{\mathbf{i}} + 150\hat{\mathbf{j}}$ km. It is sailing West, so its velocity is $\mathbf{v}_B = -10\hat{\mathbf{i}}$ km/h.

### The Magic of Relative Motion
Now, let's calculate the relative parameters. The initial position of B with respect to A is simply:

$$\mathbf{r}_{BA} = \mathbf{r}_B - \mathbf{r}_A = 80\hat{\mathbf{i}} + 150\hat{\mathbf{j}}$$

The relative velocity of B with respect to A is where the real trick happens. We subtract A's velocity from B's:

$$\mathbf{v}_{BA} = \mathbf{v}_B - \mathbf{v}_A = -10\hat{\mathbf{i}} - (30\hat{\mathbf{i}} + 50\hat{\mathbf{j}}) = -40\hat{\mathbf{i}} - 50\hat{\mathbf{j}}$$

In A's frame of reference, Ship B is moving along a straight line defined by this relative velocity vector.

### The Mathematics of Closest Approach
The shortest distance from a point (Ship A at the origin) to a line (Ship B's relative path) is the perpendicular dropped from the point to the line. The time it takes for Ship B to reach this point of closest approach can be found using a beautiful vector projection formula:

$$t = -\frac{\mathbf{r}_{BA} \cdot \mathbf{v}_{BA}}{|\mathbf{v}_{BA}|^2}$$

Why does this work? The dot product $\mathbf{r}_{BA} \cdot \mathbf{v}_{BA}$ gives us the projection of the initial position onto the velocity vector. The negative sign accounts for the fact that the ships are moving *towards* their closest point, not away from it.

### Executing the Dot Product
Let's crunch the numbers. First, the numerator (the dot product):

$$\mathbf{r}_{BA} \cdot \mathbf{v}_{BA} = (80)(-40) + (150)(-50)$$

$$\mathbf{r}_{BA} \cdot \mathbf{v}_{BA} = -3200 - 7500 = -10700$$

Next, the denominator (the square of the magnitude of relative velocity):

$$|\mathbf{v}_{BA}|^2 = (-40)^2 + (-50)^2 = 1600 + 2500 = 4100$$

### The Final Countdown
Finally, we substitute these values back into our time formula. Notice how the negative signs perfectly cancel out, ensuring we get a positive time:

$$t = -\frac{-10700}{4100} = \frac{107}{41} \approx 2.6 	ext{ hours}$$

And there we have it! In approximately 2.6 hours, the two ships will be at their absolute minimum distance from each other.

**Bonus Challenge:** If you want to find the actual minimum distance, you can use the cross product method: $d_{	ext{min}} = \frac{|\mathbf{r}_{BA} 	imes \mathbf{v}_{BA}|}{|\mathbf{v}_{BA}|}$. Try calculating it yourself to truly master the geometry of relative motion!

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Visualized Solution

$V i s u a l i z in g t h e S e t u p$

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

$R e l a t i v e P os i t i o n an d V e l oc i t y$

$T im e o f C l oses t A pp r o a c h F or m u l a$

$C a l c u l a t in g t h e D o t P r o d u c t$

$M a g ni t u d e S q u a r e d o f R e l a t i v e V e l oc i t y$

$F ina l C a l c u l a t i o n$

$F oo d f or T h o ug h t$

The Sigma Insight: Relative Velocity

The Illusion of Two Moving Bodies Imagine standing on the deck of a ship, looking out at the vast, open ocean

Setting the Stage Let's define our initial conditions

The Magic of Relative Motion Now, let's calculate the relative parameters

The Mathematics of Closest Approach The shortest distance from a point (Ship A at the origin) to a line (Ship B's relative path) is the perpendicular dropped from the point to the line

Executing the Dot Product Let's crunch the numbers

The Final Countdown Finally, we substitute these values back into our time formula

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Comprehension Passage

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Visualized Solution

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

.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; } Relative Position and Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time of Closest Approach Formula

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

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

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

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