JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2019
- Hard
- MCQ
Ship \(A\) is sailing towards north -east with velocity \(\vec v = 30\,\hat i + 50\hat j\,km/hr\) 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 \(10\, km/hr\). \(A\) will be at minimum distance from \(B\) is.........\(hrs\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(2.6\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
if\,we\,take\,the\,position\,of\,ship\,'A'\,as\,\\
origin\,then\,positions\,velocity\,of\,both\\
ships\,can\,be\,given\,as:\\
{{\vec v}_A} = \left( {30\hat i + 50\hat j} \right)km/hr\\
{{\vec v}_B} = - 10\hat i\,km/hr\,\,;\,{{\vec r}_A} = 0\hat i + \hat j\\
{{\vec r}_B} = \left( {80\hat i + 150\hat j} \right)\,km\\
Time\,after\,which\,{\rm{distance}}\,between\\
them\,will\,be\,\min imum
\end{array}\)
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} \right)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} \right)\,km\\
Time\,after\,which\,{\rm{distance}}\,between\\
them\,will\,be\,\min imum
\end{array}\)
\(\begin{array}{l}
t = \frac{{{{\vec r}_{BA}} \cdot {{\vec v}_{BA}}}}{{\left| {{{\vec v}_{BA}}} \right|}}\,;\\
Where\,{{\vec r}_{BA}} = \left( {80\hat i + 150\hat j} \right)\,km\\
{{\vec v}_{BA}} = - 10\left( {30\hat i + 50\hat j} \right)\,\left( { - 40\hat i - 50\hat j} \right)km/hr\\
\therefore \,t = - \frac{{\left( {80\hat i + 150\hat j} \right)\,.\left( { - 40\hat i - 50\hat h} \right)}}{{{{\left| { - 40i - 50\hat h} \right|}^2}}}\\
\,\,\,\,\,\, = \,\frac{{3200 + 7500}}{{4100}}hr = \frac{{10700}}{{4100}}hr = 2.6\,hrs
\end{array}\)
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