Answer
$2.96\;m/s$
Work Step by Step
The flow around the barge is uniform. Therefore, we will use the equation of continuity to solve the problem.
Let, $ A_i$ is the cross-sectional area of the canal
$v_i$ is the initial water speed
$ A_a$ is the effective cross-sectional area of the canal at a
$v_a$ is the water speed at a
$A_iv_i=A_av_a$
or, $HDv_i=[D(H-h)-d(b-h)]v_a$
or, $v_a=\frac{HDv_i}{D(H-h)-d(b-h)}$
Substituting the given values
$v_a=\frac{14\times55\times1.5}{55\times(14-0.80)-30\times(12-0.80)}\;m/s$
or, $v_a=2.96\;m/s$
Therefore, the water speed at a is $2.96\;m/s$
