After time t at position B′A′(l−x)2+(3t)2‌‌=l2 (1.5−x)2+9t2‌‌=1.52 2.25+x2−3x+9t2‌‌=2.25 x2−3x+9t2‌‌=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) Also, the rod is moving with constant velocity. So from the equation of motion, x=ut+‌
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at2 x=3t+0 x=3t Substitute 3t for x in equation (I) (3t)2−3(3t)+9t2=0 18t2−9t=0 9t(2t−1)=0 Therefore, t=‌