Answer
1.4 m
Work Step by Step
If $v_{1}>v_{2}$,
Then, $\frac{L_{1}}{t_{1}}>\frac{L_{2}}{t_{2}}$
That is, $\frac{1000\,m}{147.95\,s}>\frac{1000\,m+(L_{2}-L_{1})}{148.15\,s}$
$\implies 1000\,m\times148.15\,s>1000\,m\times147.95\,s+(L_{2}-L_{1})147.95\,s$
$\implies 200\,ms>(L_{2}-L_{1})\times147.95\,s$
$\implies 1.4\,m>(L_{2}-L_{1})$
Or $(L_{2}-L_{1})<1.4\,m$
