June 8, 2008

Q. 1.2 -- "Problems in General Physics" by I.E. Irodov

Question: A point traversed half the distance with a velocity $v_0$. 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. Find the mean velocity of the point averaged over the whole time of motion.

Answer: Let the total distance be $d$. Then, the first half was traveled in time $d/2v_0$.
After this, the remaining distance is $d/2$.
Let this distance be covered in total of time of $t$.
For $t/2$, the point traversed at speed $v_1$so the distance covered in this period is $v_1 t/2$.
The remaining distance was covered at velocity $v_2$and the distance covered at this speed is $v_2 t/2$.
We have,

\begin{displaymath}\frac{(v_1 + v_2)t}{2} = \frac{d}{2}\end{displaymath}

Which leads to:
\begin{displaymath}t = \frac {d}{v_1 + v_2} \end{displaymath}

Thus, total time of travel is:

\begin{displaymath}\frac{d}{2v_0} + \frac {d}{v_1 + v_2} \end{displaymath}

And the mean velocity over the whole time of travel is then:

\begin{displaymath}\frac{d}{\frac{d}{2v_0} + \frac {d}{v_1 + v_2}} \end{displaymath}

\begin{displaymath}= \frac{2v_0 (v_1 + v_2)}{2v_0 + v_1 + v_2} \end{displaymath}

Posted by Apurva Sharan at 2:58 PM
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