JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2016, 2022
- Medium
- MCQ
Two cars \(P\) and \(Q\) start from a point at the same time in a straight line and their positions are represented by \( x_p(t)=at+bt^2 \) and \(x_Q(t)= ft- t^2\). At what time do the cars have the same velocity \(?\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\;\frac{{f - a}}{{2\left( {1 + b} \right)}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\,\,\,Position\,of\,car\,P\,at\,any\,time\,t,\,is\\
\,\,{x_p}\left( t \right) = at + b{t^2}\\
\,\,{v_p}\left( t \right) = \frac{{d{x_p}\left( t \right)}}{{dt}} = a + 2bt\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)\\
{\rm{Similarly}},\,for\,car\,Q,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,{x_Q}\left( t \right) = ft - {t^2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,{v_Q}\left( t \right) = \frac{{d{x_q}\left( t \right)}}{{dt}} = f - 2t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)\\
\,\,\,\,\,\,\,\,\,\,{v_p}\left( t \right) = {v_Q}\left( t \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {Given} \right)\\
\therefore \,\,\,\,\,\,\,\,\,\,\,a + 2bt = f - 2t\,or,\,2t\left( {b + 1} \right) = f - a\\
\therefore \,\,\,\,\,\,\,\,\,\,\,\,t = \frac{{f - a}}{{2\left( {1 + b} \right)}}\,
\end{array}\)
\,\,\,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}\)
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