According to question.


Given,
$A B$ is an iron rod whose length is given by
$$
l=1.5 \mathrm{~m}
$$
velocity of end $\mathrm{A}$,
$$
v=3 \mathrm{~m} / \mathrm{s}
$$

Let $t$ be the time when both the ends have same speed. Since, end $A$ has always a constant velocity, therefore the velocity of end $B$ at point $B^{\prime}$ is also become $3 \mathrm{~m} / \mathrm{s}$.
In the position of $B^{\prime} A^{\prime}$ of the rod, i.e, after time $t$.
$$
\begin{aligned}
& (l-x)^2+(3 t)^2=l^2 \\
& (1.5-x)^2+9 t^2=(1.5)^2 \\
& 2.25+x^2-3 x+9 t^2=2.25 \\
& x^2-3 x+9 t^2=0
\end{aligned}
$$
As rod is moving with a constant velocity then from equation of the motion
$$
\begin{aligned}
& x=u t+\frac{1}{2} a t^2 \\
& x=3 t
\end{aligned}
$$
[ $\because$ velocity $u=3 \mathrm{~m} / \mathrm{s}$ and accleration $a=0$ ]
From Eqs. (i) and (ii)
$$
\begin{aligned}
& (3 t)^2-3(3 t)+9 t^2=0 \\
& 9 t^2-9 t+9 t^2=0 \\
& 18 t^2-9 t=0 \Rightarrow t=\frac{1}{2}
\end{aligned}
$$