A body travelling along a straight line traversed one third of the total distance with a velocity $4\text{m/s}$. The remaining part of the distance was covered with a velocity $2\text{m/s}$ for half the time and with velocity $6\text{m/s}$ for the other half of time. The mean velocity averaged over the whole time of motion is

The correct option is B $4\text{m/s}$
Let the total distance travelled by the body $=s$
Let the time taken to cover first one-third of the total distance $=t_1$
Velocity with which the body travelled first one-third of the total distance $=4\text{m/s}$
Then,
$t_1=\frac{s/3}{4}=\frac{s}{12}$


Let $t$ be the time taken to travel remaining part of the total distance.

Remaining part of the total distance $=s-\frac{s}{3}=\frac{2s}{3}$

Given that $\frac{2s}{3}$ was covered with a velocity $2\text{m/s}$ for half the time and with velocity $6\text{m/s}$ for the other half of time

$\Rightarrow \frac{2s}{3}=2\times \frac{t}{2}+6\times \frac{t}{2}=4t$