જહાજ A એ ઉત્તર-પૂર્વ દિશામાં vec v = 30,hat i | vedclass

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Question

$\begin{array}{l}
if\,we\,take\,the\,position\,of\,ship\,'A'\,as\,\
origin\,then\,positions\,velocity\,of\,both\
ships\,can\,be\,given\,a

Vedclass · Accepted Answer

$2.6$

Vedclass · Answer

$\begin{array}{l}
if\,we\,take\,the\,position\,of\,ship\,'A'\,as\,\
origin\,then\,positions\,velocity\,of\,both\
ships\,can\,be\,given\,as:\
{{\vec v}_A} = \left( {30\hat i + 50\hat j} ight)km/hr\
{{\vec v}_B} =  - 10\hat i\,km/hr\,\,;\,{{\vec r}_A} = 0\hat i + \hat j\
{{\vec r}_B} = \left( {80\hat i + 150\hat j} ight)\,km\
Time\,after\,which\,{m{distance}}\,between\
them\,will\,be\,\min imum
\end{array}$

$\begin{array}{l}
t = \frac{{{{\vec r}_{BA}} \cdot {{\vec v}_{BA}}}}{{\left| {{{\vec v}_{BA}}} ight|}}\,;\
Where\,{{\vec r}_{BA}} = \left( {80\hat i + 150\hat j} ight)\,km\
{{\vec v}_{BA}} =  - 10\left( {30\hat i + 50\hat j} ight)\,\left( { - 40\hat i - 50\hat j} ight)km/hr\
	herefore \,t =  - \frac{{\left( {80\hat i + 150\hat j} ight)\,.\left( { - 40\hat i - 50\hat h} ight)}}{{{{\left| { - 40i - 50\hat h} ight|}^2}}}\
\,\,\,\,\,\, = \,\frac{{3200 + 7500}}{{4100}}hr = \frac{{10700}}{{4100}}hr = 2.6\,hrs
\end{array}$

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