Question 115: Motion in 2D & Vectors

Question

Ship A is sailing towards north-east with
velocity $$\mathop v\limits^ 	o   = 30\mathop i\limits^ \wedge   + 50\mathop j\limits^ \wedge  $$ km/hr where $$\mathop i\limits^ \wedge  $$ points
east and $$\mathop j\limits^ \wedge  $$ , 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/hr. A will be
at minimum distance from B in :

Accepted Answer

Considering the initial position of ship A as origin, so the velocity and position of ship will be
$${\overrightarrow v _A} = (30\widehat i + 50\widehat j)$$ and $${\overrightarrow r _A} = (0\widehat i + 0\widehat j)$$
Now, as given in the question, velocity and position of ship B will be, $${\overrightarrow v _B} =  - 10\widehat i$$ and $${\overrightarrow r _B} = (80\widehat i + 150\widehat j)$$
Time after which the distance is minimum between A and B can be calculated as
$$t = {{|{{\overrightarrow r }_{BA}}.\,{{\overrightarrow v }_{BA}}|} \over {|{{\overrightarrow v }_{BA}}{|^2}}}$$
where, $${\overrightarrow r _{BA}} = {\overrightarrow r _B} - {\overrightarrow r _A} = 80\widehat i + 150\widehat j$$
and $${\overrightarrow v _{BA}} =  - 10\widehat i - (30\widehat i + 50\widehat j)$$
$$ =  - 40\widehat i - 50\widehat j$$
$$ \Rightarrow t = {{|(80\widehat i + 150\widehat j)\,.\,( - 40\widehat i - 50\widehat j)|} \over {| - 40\widehat i - 50\widehat j{|^2}}}$$
$$ = {{3200 + 7500} \over {4100}} = {{10700} \over {4100}} = 2.6$$ h

Answer

2.2 hrs

Answer

4.2 hrs

Answer

2.6 hrs

Answer

3.2 hrs

Motion in 2D & Vectors

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