v2=2f(d+n)=2f′dv=f′(t)=(m+t)f eliminate d and m we get (f′−f)n=21ff′m2.
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
Held on 30 Apr 2005 · Verified 6 Jul 2026.
(f−f′)m2=ff′n
(f+f′)m2=ff′n
21(f+f′)m=ff′n2
(f′−f)n=21ff′m2
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