JEE Mains · Physics · STD 11 - 2. motion in straight line
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
- A \(\frac{{2{a_1}{a_2}}}{{{a_1} + {a_2}}}t\)
- B \(\sqrt {2{a_1}{a_2}}\,t\)
- C \(\sqrt {{a_1}{a_2}}\,t\)
- D \(\frac{{{a_1} + {a_2}}}{2}t\)
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 } =…
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