Question

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

• A. $5\text{m/s}$
• B. $4\text{m/s}$
• C. $4.5\text{m/s}$
• D. $3.5\text{m/s}$

Answer 1

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

$\Rightarrow \frac{2s}{3}=2\times \frac{t}{2}+6\times \frac{t}{2}=4t$
$\therefore t=\frac{2s}{12}=\frac{s}{6}$

$\therefore \text{Average velocity}=\frac{\text{total displacement}}{\text{total time}}$
$=\frac{s}{t_1+t}=\frac{s}{\frac{s}{12}+\frac{s}{6}}=\frac{12\times 6}{12+6}=4\text{m/s}$