Question

A particle covers each one third of Total distance with speed V1, V2 and V3 respectively. find the average speed of the particle?

Answer 1

Total distance traveled = S 
Time taken for the 3 parts of the journey = t1, t2 and t3 respectively 
Speed during the 3 parts of the journey = v1, v2, v3, respectively 
Distance covered in part of the journey = S/3 
=> t1 = (S/3)/v1 = S/(3v1) 
t2 = (S/3)/v2 = S/(3v2) 
t3 = (S/3)/v3 = S/(3v3) 
Total time taken = t = t1 + t2 + t3 
= (S/3)(1/v1 + 1/v2 + 1/v3) 
Average speed = V(av) = S/t = S/[(S/3)(1/v1 + 1/v2 + 1/v3)] 
= 1/[(1/3)(1/v1 + 1/v2 + 1/v3)] 
= 1/[(v2 v3 + v3 v1 + v1v2)/{3(v1 v2 v3)}] 
= (3 v1 v2 v3) / (v1 v2 + v2 v3 + v3 v1

Answer 2

Answer:

Concept:

The overall distance the object covers in a given amount of time is its average speed. A scalar value represents the average speed. It has no direction and is indicated by the magnitude.

Explanation:

Average speed = total distance covered/total time taken

let the total distance  $= 3 x$

Time is taken to cover the first one third  $(x) = t_{1}=\frac{x}{v_{1}}$

Time is taken to cover the second one third $(x)= t_{2}=\frac{x}{v_{2}}$

Time is taken to cover the third one third  $(x)=t_{3}=\frac{x}{v_{3}}$

                 $\begin{aligned}\text { average speed } &=\frac{3 x}{\frac{x}{v_{1}}+\frac{x}{v_{2}}+\frac{x}{v_{3}}} \\\\&=\frac{3 x}{x\left(\frac{v_{3} v_{2}+v_{1} v_{3}+v_{1} v_{2}}{v_{1} v_{2} v_{3}}\right)}\end{aligned}$

average speed = ${3 v_{1} v_{2} v_{3}}$ / $v_{3} v_{2}+v_{1} v_{3}+v_{1} v_{2}$

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