JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2019
- Hard
- MCQ
In a car race on straight road, car \(A\) takes a time \(t\) less than car \(B\) at the finish and passes finishing point with a speed \(v\) more than that of car \(B.\) Both the cars start from rest and travel with constant acceleration \(a_1\) and \(a_2\) respectively. Then \(v\) is equal to
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\sqrt {{a_1}{a_2}}\,t\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\sqrt {\frac{{2\ell }}{{{a_2}}}} - \sqrt {\frac{{2\ell }}{{{a_1}}}} = t\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\frac{{\sqrt {2\ell } }}{t} = \frac{{\sqrt {{a_1}{a_2}} }}{{{{\sqrt a }_1} - \sqrt {{a_2}} }}\\
\sqrt {2{a_1}\ell } - \sqrt {2{a_2}\ell } = v\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\frac{{\sqrt {2\ell } }}{v} = \frac{1}{{\sqrt {{a_1}} - \sqrt {{a_2}} }}\\
\Rightarrow \,\,\frac{v}{t} = \sqrt {{a_1}{a_2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,v = \left( {\sqrt {{a_1}{a_2}} \,\,\,} \right)t
\end{array}\)
\sqrt {\frac{{2\ell }}{{{a_2}}}} - \sqrt {\frac{{2\ell }}{{{a_1}}}} = t\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\frac{{\sqrt {2\ell } }}{t} = \frac{{\sqrt {{a_1}{a_2}} }}{{{{\sqrt a }_1} - \sqrt {{a_2}} }}\\
\sqrt {2{a_1}\ell } - \sqrt {2{a_2}\ell } = v\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\frac{{\sqrt {2\ell } }}{v} = \frac{1}{{\sqrt {{a_1}} - \sqrt {{a_2}} }}\\
\Rightarrow \,\,\frac{v}{t} = \sqrt {{a_1}{a_2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,v = \left( {\sqrt {{a_1}{a_2}} \,\,\,} \right)t
\end{array}\)
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