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दो कारें $P$ तथा $Q$ एक ही समय पर किसी बिन्दु से एक सरल रेखा में चलना प्रारम्भ करती हैं तथा उनकी स्थितियों को क्रमश: $ x_p(t)=at+bt^2 $ तथा $x_Q(t)= ft- t^2$ से निरूपित किया जाता है । किस समय पर इन दोनों का वेग समान होगा ?
$\frac{{a + f}}{{2\left( {1 + b} \right)}}$
$\;\frac{{f - a}}{{2\left( {1 + b} \right)}}$
$\;\frac{{a + f}}{{\left( {1 + b} \right)}}$
$\;\frac{{a + f}}{{2\left( {b - 1} \right)}}$
Solution
$\begin{array}{l}
\,\,\,Position\,of\,car\,P\,at\,any\,time\,t,\,is\\
\,\,{x_p}\left( t \right) = at + b{t^2}\\
\,\,{v_p}\left( t \right) = \frac{{d{x_p}\left( t \right)}}{{dt}} = a + 2bt\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( i \right)\\
{\rm{Similarly}},\,for\,car\,Q,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,{x_Q}\left( t \right) = ft – {t^2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,{v_Q}\left( t \right) = \frac{{d{x_q}\left( t \right)}}{{dt}} = f – 2t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,…\left( {ii} \right)\\
\,\,\,\,\,\,\,\,\,\,{v_p}\left( t \right) = {v_Q}\left( t \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {Given} \right)\\
\therefore \,\,\,\,\,\,\,\,\,\,\,a + 2bt = f – 2t\,or,\,2t\left( {b + 1} \right) = f – a\\
\therefore \,\,\,\,\,\,\,\,\,\,\,\,t = \frac{{f – a}}{{2\left( {1 + b} \right)}}\,
\end{array}$
