Ship `A` is sailing towards north-east with velocity `vecv=30hati+50hatJ km//hr` where `hati` points east and `hatj`, north. Ship `B` is at a distance of `80km` east and `150km` north of Ship `A` and is sailing towards west at `10km//hr`. A will be at minimum distance from `B` in:

Options

A. `4.2` hrs.
B. `2.6` hrs.
C. `3.2` hrs.
D. `2.2`hrs.

Answer

To solve the problem step by step, we will analyze the motion of both ships and find the time at which the distance between them is minimized.

### Step 1: Define the positions and velocities of the ships
- Ship A is at the origin (0, 0) and has a velocity vector:
\[ \vec{v_A} = 30 \hat{i} + 50 \hat{j} \text{ km/hr} \]
- Ship B is at the position (80, 150) km and is moving towards the west with a velocity vector:
\[ \vec{v_B} = -10 \hat{i} \text{ km/hr} \]

### Step 2: Write the position vectors of both ships as functions of time
Let \( t \) be the time in hours. The position of Ship A as a function of time is:
\[ \vec{r_A}(t) = (30t) \hat{i} + (50t) \hat{j} \]
The position of Ship B as a function of time is:
\[ \vec{r_B}(t) = (80 - 10t) \hat{i} + (150) \hat{j} \]

### Step 3: Find the relative position vector between Ship B and Ship A
The relative position vector from Ship A to Ship B is given by:
\[ \vec{r_{AB}}(t) = \vec{r_B}(t) - \vec{r_A}(t) \]
Substituting the position vectors:
\[ \vec{r_{AB}}(t) = [(80 - 10t) - (30t)] \hat{i} + [150 - (50t)] \hat{j} \]
This simplifies to:
\[ \vec{r_{AB}}(t) = (80 - 40t) \hat{i} + (150 - 50t) \hat{j} \]

### Step 4: Find the distance squared between the two ships
The distance squared \( D^2(t) \) between the two ships is:
\[ D^2(t) = (80 - 40t)^2 + (150 - 50t)^2 \]

### Step 5: Differentiate the distance squared with respect to time
To find the minimum distance, we differentiate \( D^2(t) \) with respect to \( t \) and set the derivative to zero:
\[ \frac{dD^2}{dt} = 2(80 - 40t)(-40) + 2(150 - 50t)(-50) = 0 \]
Simplifying:
\[ -80(80 - 40t) - 100(150 - 50t) = 0 \]
Expanding and combining terms:
\[ -6400 + 3200t - 15000 + 5000t = 0 \]
\[ 8200t - 21400 = 0 \]
Solving for \( t \):
\[ t = \frac{21400}{8200} = 2.61 \text{ hours} \]

### Step 6: Conclusion
The time at which Ship A will be at minimum distance from Ship B is approximately:
\[ \boxed{2.61 \text{ hours}} \]