Question

A point traversed half the distance with velocity ${{v}_{o}}$. The remaining part of the distance was covered with velocity ${{v}_{1}}$ for half the time, and with velocity ${{v}_{2}}$ for the other half of the time. If the mean velocity of the point averaged over the whole time of the motion is $\left\langle v \right\rangle =\dfrac{x{{v}_{o}}({{v}_{1}}+{{v}_{2}})}{2{{v}_{o}}+{{v}_{1}}+{{v}_{2}}}$. Find $x$.

Answer 1

Hint: Since the expression for the average value of velocity is needed, calculate the total distance covered and the total time using the basic definition of speed. Apply the formula on for each part of the journey to find the total time and distance. Substitute these values in the expression for average velocity and compare to get the answer.
Formula used: Formula of speed:
$\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$
Formula for average speed:
$\left\langle v \right\rangle =\dfrac{\text{Total distance}}{\text{Total time}}$

Complete step-by-step answer:
Let us understand the question in steps.
The first part of the motion:
If the total distance covered by the point is taken as ${{D}_{0}}$. Then in this part, the distance was half of it. That is $\dfrac{{{D}_{0}}}{2}$ .
The velocity of the point throughout this part is uniform and is ${{v}_{1}}$.
Let the time taken by the point to travel this part of the motion be ${{t}_{1}}$. Then, by applying the formula of velocity. We get:
$\begin{aligned} & \text{Speed}=\dfrac{\text{Distance}}{\text{Time}} \\\ & \Rightarrow \text{Time}=\dfrac{\text{Distance}}{\text{Speed}} \\\ & \Rightarrow {{t}_{1}}=\dfrac{{{D}_{0}}}{2{{v}_{1}}} \\\ \end{aligned}$
The second part of the motion:
Let it take a total time of ${{t}_{2}}$. Where its further parts take half the time each that is, each interval takes $\dfrac{{{t}_{2}}}{2}$time.
It is given that the uniform velocities for these time intervals are ${{v}_{2}}$ and ${{v}_{3}}$ respectively.
When velocity was ${{v}_{2}}$, the distance covered equals:

& \text{Distance}=\text{Time}\times \text{Speed} \\\ & \Rightarrow {{d}_{2}}=\dfrac{{{t}_{2}}{{v}_{2}}}{2} \\\ \end{aligned}$$ Similarly, when velocity was ${{v}_{3}}$: $$\begin{aligned} & \text{Distance}=\text{Time}\times \text{Speed} \\\ & \Rightarrow {{d}_{3}}=\dfrac{{{t}_{2}}{{v}_{3}}}{2} \\\ \end{aligned}$$ Since, the distance covered in the second part was $\dfrac{{{D}_{0}}}{2}$, $$\begin{aligned} & \dfrac{{{D}_{0}}}{2}={{d}_{2}}+{{d}_{3}} \\\ & \Rightarrow \dfrac{{{D}_{0}}}{2}=\dfrac{{{t}_{2}}{{v}_{2}}}{2}+\dfrac{{{t}_{2}}{{v}_{3}}}{2} \\\ & \Rightarrow {{D}_{0}}={{t}_{2}}({{v}_{2}}+{{v}_{3}}) \\\ & \Rightarrow {{t}_{2}}=\dfrac{{{D}_{0}}}{({{v}_{2}}+{{v}_{3}})} \\\ \end{aligned}$$ The average velocity for a motion is the ratio of the total distance travelled to the total time. $$\begin{aligned} & \left\langle v \right\rangle =\dfrac{\text{Total distance}}{\text{Total time}} \\\ & \left\langle v \right\rangle =\dfrac{{{\text{D}}_{0}}}{{{\text{t}}_{1}}+{{t}_{2}}}=\dfrac{{{\text{D}}_{0}}}{\dfrac{{{D}_{0}}}{2{{v}_{1}}}+\dfrac{{{D}_{0}}}{({{v}_{2}}+{{v}_{3}})}} \\\ & \left\langle v \right\rangle =\dfrac{2{{v}_{1}}({{v}_{2}}+{{v}_{3}})}{2{{v}_{1}}+{{v}_{2}}+{{v}_{3}}} \\\ \end{aligned}$$ Comparing this with the given equation we get the value of $$x=2$$ . Therefore, the answer to this question is $$x=2$$ . Note: The motion of the point is in one dimension and it moves in only one direction. Therefore, it is alright to use the formulae with speed and distance rather than velocity and displacement. As essentially, the magnitude of velocity is speed and similarly, the magnitude of displacement is distance.