1. Set up the coordinate system

Let east be the $x$-axis and north be the $y$-axis.

Velocity of ship A: $\vec v_A = 30\hat i + 50\hat j \quad \text{km/hr}$

Ship B is initially $80$ km east and $150$ km north of A, so the initial position of B relative to A is $\vec r_0 = 80\hat i + 150\hat j$

Ship B is sailing towards west at $10$ km/hr, so $\vec v_B = -10\hat i$

2. Find relative velocity of B with respect to A

$\vec v_{BA} = \vec v_B - \vec v_A$ $\vec v_{BA} = (-10\hat i) - (30\hat i + 50\hat j)$ $\vec v_{BA} = -40\hat i - 50\hat j$

So the relative position after time $t$ is $\vec r(t) = \vec r_0 + \vec v_{BA} t$ $\vec r(t) = (80-40t)\hat i + (150-50t)\hat j$

3. Condition for minimum distance

Minimum distance occurs when $|\vec r(t)|$ is minimum. It is easier to minimize its square: $D^2(t) = (80-40t)^2 + (150-50t)^2$

Differentiate with respect to $t$ and set equal to zero: $\frac{d}{dt}D^2(t) = 2(80-40t)(-40) + 2(150-50t)(-50)=0$

Divide by $2$: $-40(80-40t) - 50(150-50t)=0$

$-3200 + 1600t - 7500 + 2500t = 0$

$4100t - 10700 = 0$

$t = \frac{10700}{4100} = \frac{107}{41} \approx 2.61\text{ hr}$

4. Match with the options

$t \approx 2.6\text{ hr}$

So the correct option is C.

5. Verification with stored answer

Stored correct answer: C

Our derived answer: C

Hence, they agree.