10 ,km,h^{-1} ઝડપે પશ્વિમ દિશામાં એક વહાણ A ગ | vedclass

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Question

$\begin{array}{l}
Given\,situation\,is\,shown\,in\,the\,figure.\
Velocity\,of\,ship\,A,\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{v_A} = 10\,km

Vedclass · Accepted Answer

$5$

Vedclass · Answer

$\begin{array}{l}
Given\,situation\,is\,shown\,in\,the\,figure.\
Velocity\,of\,ship\,A,\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{v_A} = 10\,km\,{h^{ - 1}}\,towards\,west\
Velocity\,of\,ship\,B,\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{v_B} = 10\,km\,{h^{ - 1}}\,towards\,north\
\,\,\,\,\,\,\,\,\,\,OS = 100\,km\
\,\,\,\,\,\,\,\,\,OP = \,shortest\,{m{distance}}\
{m{Relative}}\,{m{velocity}}\,{m{between}}\,{m{A}}\,{m{and}}\,{m{B}}\,{m{is}}\
\,\,\,\,\,\,\,\,\,{{m{v}}_{AB}} = \sqrt {v_A^2 + v_B^2}  = 10\sqrt 2 \,km\,{h^{ - 1}}
\end{array}$

$\begin{array}{l}
\cos {45^ \circ } = \frac{{OP}}{{OS}};\frac{1}{{\sqrt 2 }} = \frac{{OP}}{{100}}\
\,\,\,OP = \frac{{100}}{{\sqrt 2 }} = \frac{{100\sqrt 2 }}{2} = 50\sqrt 2 \,km\
The\,time\,after\,which\,{m{distance}}\,{m{between}}\,{m{them}}\
{m{equals}}\,{m{to}}\,{m{OP}}\,is\,given\,by\
\,\,\,\,\,t = \frac{{OP}}{{{v_{AB}}}} = \frac{{50\sqrt 2 }}{{10\sqrt 2 }} \Rightarrow t = 5h
\end{array}$

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