Question 28: Motion in 1D

Question

A person travels $$x$$ distance with velocity $$v_{1}$$ and then $$x$$ distance with velocity $$v_{2}$$ in the same direction. The average velocity of the person is $$\mathrm{v}$$, then the relation between $$v, v_{1}$$ and $$v_{2}$$ will be.

Accepted Answer

The average velocity is defined as the total displacement divided by the total time. Here, the person travels the same distance $x$ twice, once with velocity $v_1$ and once with velocity $v_2$.
The time to travel distance $x$ with velocity $v_1$ is $t_1 = \frac{x}{v_1}$, and the time to travel distance $x$ with velocity $v_2$ is $t_2 = \frac{x}{v_2}$.
The total time is then
$t = t_1 + t_2 = \frac{x}{v_1} + \frac{x}{v_2}$.
The total displacement is $2x$. So, the average velocity $v$ is given by
$ v = \frac{	ext{total displacement}}{	ext{total time}} = \frac{2x}{\frac{x}{v_1} + \frac{x}{v_2}} = \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}} $
Multiplying both sides by $2$, we get
$ \frac{2}{v} = \frac{1}{v_1} + \frac{1}{v_2} $

Answer

$$\mathbf{V}=\mathbf{V}_{1}+\mathbf{V}_{2}$$

Answer

$$V=\frac{v_{1}+V_{2}}{2}$$

Answer

$$\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{v}_{1}}+\frac{1}{\mathrm{v}_{2}}$$

Answer

$$\frac{2}{\mathrm{~V}}=\frac{1}{\mathrm{v}_{1}}+\frac{1}{\mathrm{v}_{2}}$$

Motion in 1D

A person travels $x distance with velocity v_{1} and then x distance with velocity v_{2} in the same direction. The average velocity of the person is \mathrm{v}, then the relation between v, v_{1} and v_{2}$ will be.

Choose the correct answer: