A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is:

Let the total distance covered by the object is 'X' meter.
Hence, the distance travelled in first half is $\frac{X}{2}$
Time taken by the object in the first half = $t_1= \frac{{\displaystyle \raisebox{1ex}{$X$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}$
$t_1 = \frac{X}{6}$

Time taken by the object in the second half = $t_2 = t_3 = \frac{X}{24}$

$V_{avg} = \frac{X}{\left({\displaystyle \frac{X}{6}} +{\displaystyle \frac{X}{24}} + {\displaystyle \frac{X}{24}}\right)}$

$V_{avg} = \left[\frac{X}{\left({\displaystyle \raisebox{1ex}{$X$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}\right]$

$V_{avg} = 4 m/s$