Two points A and B move from rest along a straight line with constant acceleration f and f' respectively. If A takes m sec. more than B and describes - Mathematics MCQ AIEEE 2026 | ExamTest.live

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Two points A and B move from rest along a straight line with constant acceleration f and f^{'} respectively. If A takes m sec. more than B and describes n units more than B in acquiring the same speed then:

Solution & Step-by-step Explanation

Let common speed be v .For A : v = f t_{A} ⟹ t_{A} = v / f . s_{A} = v^{2} / (2 f) .For B : v = f^{'} t_{B} ⟹ t_{B} = v / f^{'} . s_{B} = v^{2} / (2 f^{'}) .Given t_{A} - t_{B} = m : v (1/ f - 1/ f^{'}) = m ⟹ v (f^{'} - f) = f f^{'} m ⟹ v = \frac{f f ^{'} m}{f ^{'} - f} .Given s_{A} - s_{B} = n : \frac{v ^{2}}{2} (1/ f - 1/ f^{'}) = n ⟹ \frac{v ^{2}}{2} \frac{f ^{'} - f}{f f ^{'}} = n Substitute v : \frac{1}{2} (\frac{f f ^{'} m}{f ^{'} - f})^{2} \frac{f ^{'} - f}{f f ^{'}} = n \frac{1}{2} \frac{( f f ^{'} ) ^{2} m ^{2}}{( f ^{'} - f ) ^{2}} \frac{f ^{'} - f}{f f ^{'}} = n \frac{1}{2} \frac{f f ^{'} m ^{2}}{f ^{'} - f} = n ⟹ (f^{'} - f) n = \frac{1}{2} f f^{'} m^{2} .

Try it yourself before checking the explanation above.

Two points A and B move from rest along a straight line with constant acceleration f and f^{'} respectively. If A takes m sec. more than B and describes n units more than B in acquiring the same speed then:

(f - f^{'}) m^{2} = f f^{'} n

(f + f^{'}) m^{2} = f f^{'} n

\frac{1}{2} (f + f^{'}) m = f f^{'} n

(f^{'} - f) n = \frac{1}{2} f f^{'} m^{2}

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