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Step-by-Step Solution
Step 1: Identify the known quantities
• Initial position of Ship A, $ \overrightarrow{r}_A = 0\,\hat{i} + 0\,\hat{j} $.
• Velocity of Ship A, $ \overrightarrow{v}_A = 30\,\hat{i} + 50\,\hat{j} $ (km/hr).
• Initial position of Ship B, $ \overrightarrow{r}_B = 80\,\hat{i} + 150\,\hat{j} $.
• Velocity of Ship B, $ \overrightarrow{v}_B = -10\,\hat{i} $ (km/hr, meaning it sails west).
Step 2: Define relative position and relative velocity
• The relative position of B with respect to A at $ t=0 $ is
$$
\overrightarrow{r}_{BA}
= \overrightarrow{r}_B - \overrightarrow{r}_A
= (80\,\hat{i} + 150\,\hat{j}) - (0\,\hat{i} + 0\,\hat{j})
= 80\,\hat{i} + 150\,\hat{j}.
$$
• The relative velocity of B with respect to A is
$$
\overrightarrow{v}_{BA}
= \overrightarrow{v}_B - \overrightarrow{v}_A
= \bigl(-10\,\hat{i}\bigr) - \bigl(30\,\hat{i} + 50\,\hat{j}\bigr)
= -10\,\hat{i} - 30\,\hat{i} - 50\,\hat{j}
= -40\,\hat{i} - 50\,\hat{j}.
$$
Step 3: Use the formula for time at which distance is minimum
The time $ t $ at which the distance between the two ships is minimum is given by
$$
t
= \frac{\bigl|\overrightarrow{r}_{BA}\cdot\overrightarrow{v}_{BA}\bigr|}{\bigl|\overrightarrow{v}_{BA}\bigr|^2}.
$$
Step 4: Calculate the dot products and magnitudes
1. Dot product $ \overrightarrow{r}_{BA}\cdot \overrightarrow{v}_{BA} $:
$$
\overrightarrow{r}_{BA}
= 80\,\hat{i} + 150\,\hat{j}, \quad
\overrightarrow{v}_{BA}
= -40\,\hat{i} - 50\,\hat{j}.
$$
Hence,
$$
\overrightarrow{r}_{BA}\cdot \overrightarrow{v}_{BA}
= (80)(-40) + (150)(-50)
= -3200 - 7500
= -10700.
$$
The absolute value is $ | -10700 | = 10700 $.
2. Magnitude $ |\overrightarrow{v}_{BA}|^2 $:
$$
|\overrightarrow{v}_{BA}|^2
= (-40)^2 + (-50)^2
= 1600 + 2500
= 4100.
$$
Step 5: Compute the time $ t $
Substitute these values into the formula:
$$
t
= \frac{10700}{4100}
= 2.6 \text{ hours}.
$$
Step 6: Conclude the correct answer
Therefore, the time after which Ship A is at the minimum distance from Ship B is 2.6 hours.
