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Question:

Ship A is sailing towards north-east with velocity (\vec{v} = 30\hat{i} + 50\hat{j}) km/h where (\hat{i}) points east and (\hat{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. At what time will Ship A be at minimum distance from Ship B?

4.2 hrs.

3.2 hrs.

2.6 hrs.

2.2 hrs.

Solution:

Correct option is D. 2.6 hrs.
If we take the position of ship A as origin then position and velocities of both ships can be given as:
(\vec{V}A = (30\hat{i} + 50\hat{j})) km/h
(\vec{V}B = -10\hat{i}) km/h
(\vec{r}A = 0\hat{i} + 0\hat{j})
(\vec{r}B = (80\hat{i} + 150\hat{j})) km
Time after which distance between them will be minimum
(t = \frac{-\vec{r}
{BA} \cdot \vec{V}
{BA}}{|\vec{V}
{BA}|^2})
where
(\vec{r}
{BA} = (80\hat{i} + 150\hat{j})) km
(\vec{V}_{BA} = -10\hat{i} - (30\hat{i} + 50\hat{j}) = (-40\hat{i} - 50\hat{j})) km/h
(\therefore t = \frac{-(80\hat{i} + 150\hat{j}) \cdot (-40\hat{i} - 50\hat{j})}{|-40\hat{i} - 50\hat{j}|^2} = \frac{3200 + 7500}{40^2 + 50^2} hr = \frac{10700}{4100} hr = 2.6 hrs.)