Question

If a particle travels n equal distances with speeds V1​,V2​,....Vn​, then the average speed V of the particle will be such that
A) $V = \dfrac{{{v_1} + {v_2} + {v_3} + ....... + {v_n}}}{n}$.
B) $V = \dfrac{{n{v_1}{v_2} + {v_n}}}{{{v_1} + {v_2} + {v_3} + ....... + {v_n}}}$.
C) $\dfrac{1}{V} = \dfrac{1}{n}(\dfrac{1}{{{v_1}}} + \dfrac{1}{{{v_2}}} + \dfrac{1}{{{v_3}}} + ........ + \dfrac{1}{{{v_n}}})$.
D ) $V = \sqrt {v_1^2 + v_2^2 + v_3^2 + ...........v_n^2}$.

Answer 1

Hint
Use the formula distance = speed $\times$time. And take out overall time and hence the speed by computing with the formula $V = \dfrac{{Total\,dis\tan ce\,travelled}}{{Total\,time\,taken}}$ $= \dfrac{n}{{{t_1} + {t_2} + {t_3} + {t_4} + ..... + {t_n}}}$to solve the problem.

Step by step Solution
Divide the problem into 3-4 steps first because equal distances are covered with different velocities and write their time for each duration. Let the common n equal distance have a value of s metres.
${t_1} = \dfrac{s}{{{v_1}}}$ and similarly ,
$\Rightarrow {t_2} = \dfrac{s}{{{v_2}}}$
$\Rightarrow {t_3} = \dfrac{s}{{{v_3}}}$
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${t_n} = \dfrac{s}{{{v_n}}}$
Now,
The average speed in the entire journey is given by,
$\begin{gathered} V = \dfrac{{Total\,dis\tan ce\,travelled}}{{Total\,time\,taken}} \\\ v = \dfrac{n}{{{t_1} + {t_2} + {t_3} + {t_4} + ..... + {t_n}}} \\\ \end{gathered}$
Putting the values from the above solved part we have,
$v = \dfrac{n}{{\dfrac{s}{{{v_1}}} + \dfrac{s}{{{v_2}}} + \dfrac{s}{{{v_3}}} + ....... + \dfrac{s}{{{v_n}}}}}$
And finally we get ,
$v = \dfrac{n}{{\dfrac{1}{n}(\dfrac{s}{{{v_1}}} + \dfrac{s}{{{v_2}}} + \dfrac{s}{{{v_3}}} + ....... + \dfrac{s}{{{v_n}}})}}$
Hence the correct option is option C .

Note
There are other kinds of questions where the distance is covered by the first half and second half . Those questions need to be solved by taking ${d_1} + {d_2}$in the similar way like we took time in this case.
Further memorizing these equations helps to solve these kinds of problems faster.