If particled travels `n` equal distances with speeds `v_(1), v_(2),…v_(n)`, then the average speed `vec v` of the particle will be such that .

Options

A. `vecV=(v_(1) +v_(2) +.......+v_(n))/(n)`
B. `vecV=(nv_(1)v_(2)+v_(n))/(v_(1)+v_(2)+v_(3)+....+v_(n))`
C. `(1)/(barV)=(1)/(n) ((1)/(v_(1))+(1)/(v_(2))+.......+(1)/(v_(n)))`
D. `barV=sqrt(v_(1)^(2)+v_(2)^(2)+.....+(1)/(v_(n)))`

Answer

C

Explanation

`t_(1) =(S)/(v_(1)), t_(2) =(s)/(v_(2)) ,…..t_(n) =(s)/(v_(n))`
Average speed =`(Total distance )//(Total time)`
`rArr vec V =(ns)/(t_(1)(+t_(2))+….+t_(n)))`
`(ns)/((s)/(v_(1))+(s)/(v_(2)) +…+(s)/(v_(n))) =(n)/((1)/(v_(1)) +(1)/(v_(2))+…+(1)/(v_(n)))`
Taking reciprocal, we get `(1)/(bar(V))=(1)/(n)((1)/(v_(1))+........+(1)/(v_(n)))`.