Question

A person travelling in a straight line moves with a constant velocity of $10\text{m/s}$ for a distance of $100\text{m}$ and with a constant velocity of $15\text{m/s}$ for the next equal distance in the same direction. The average velocity of the motion (in $\text{m/s}$) is

Answer 1

Given,
Velocity of first half $\left(v_1\right)=10\text{m/s}$
Velocity of second half $\left(v_2\right)=15\text{m/s}$
Displacement with velocity $v_1$ $d_1=100\text{m}$
Displacement with velocity $v_2$ $d_2=100\text{m}$
Now,
Time taken for displacement $d_1$ with velocity $v_1$ is given by
$t_1=\frac{d_1}{v_1}=\frac{100}{10}=10\text{sec}...........\left(1\right)$
And,
Time taken for displacement $d_2$ with velocity $v_2$
$t_2=\frac{d_2}{v_2}=\frac{100}{15}=\frac{20}{3}\text{sec}.........\left(2\right)$
We know,
Average velocity $=\frac{\text{Total displacement}}{\text{Total time}}$
$v_{av}=\frac{d_1+d_2}{t_1+t_2}$
$\Rightarrow v_{av}=\frac{200}{\frac{50}{3}}$
$\Rightarrow v_{av}=12\text{m/s}$
The average velocity of the person travelling is $12\text{m/s}$
Hence, the answer is $12$.