Question

A body traveling along a straight line traversed first 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 average velocity over the whole time of motion is

• A. $5\text{m/s}$
• B. $4\text{m/s}$
• C. $4.5\text{m/s}$
• D. $3\text{m/s}$

Answer 1

The correct option is B $4\text{m/s}$

Let total distance $=s$
Distance covered with a velocity of $4\text{m/s}=\frac{s}{3}$
Let the time taken to cover first one third distance $=t_1$, then
$t_1=\frac{\frac{s}{3}}{4}=\frac{s}{12}$
Now let $t_2$ be the time for the rest two journeys.
Distance covered in rest of the two journeys $=s-\frac{s}{3}=\frac{2s}{3}$
Since half of the remaining distance is covered at $2m/s$ and half at $6m/s$. Therefore
$\frac{2s}{3}=2t_2+6t_2=8t_2$
$\therefore t_2=\frac{2s}{24}=\frac{s}{12}$
Total time taken to cover rest of the two journeys $=2t_2=\frac{s}{6}$
$\therefore$ Average velocity $=\frac{\text{total displacement}}{\text{total time}}$
$=\frac{s}{t_1+2t_2}=\frac{s}{\frac{s}{12}+\frac{s}{6}}=\frac{12\times 6}{12+6}=4\text{m/s}$