If a body covers three equal distances with speeds $v_{1} v_{2} a n d v_{3}$ then what is the average speed of the body?

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Solution

Step 1: Definition of used terms
Speed is defined as the rate of change of distance with respect to time is called speed.
$s p e e d = \frac{d i s \tan c e t r a v e l e d}{t i m e t a k e n}$
$t i m e t a k e n = \frac{d i s \tan c e t r a v e l e d}{s p e e d}$
Average speed: The average speed of a body is defined as the ratio of the total distance traveled by the body to the total time taken by the body.
average speed = $\frac{t o t a l d i s \tan c e}{t o t a l t i m e}$ .
Step 2: Calculation
Let d is the three equal distances covered by the body with speeds v₁, v₂, v₃ in time intervals t₁, t₂, t₃ respectively
Time is taken to complete d distance with velocity v₁ is $t_{1} = \frac{d}{v_{1}}$
Time taken to complete d distance with velocity v₂ is $t_{2} = \frac{d}{v_{2}}$
And the time taken to complete d distance with velocity v₃ is $t_{3} = \frac{d}{v_{3}}$
Now total time T is,
$T = t_{1} + t_{2} + t_{3} T = \frac{d}{v_{1}} + \frac{d}{v_{2}} + \frac{d}{v_{3}} T = d (\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}})$
Substituting the values in the equation of average speed,
$a v e r a g e s p e e d = \frac{T o t a l d i s \tan c e}{t o t a l t i m e} A v e r a g e s p e e d = \frac{d + d + d}{T}$
= $\frac{3 d}{d (\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}})}$
= $\frac{3}{\frac{1}{v_{1}} + \frac{1}{v_{2}} + \frac{1}{v_{3}}}$
$= \frac{3}{\frac{v_{3} v_{2} + v_{1} v_{3} + v_{1} v_{2}}{v_{1} v_{2} v_{3}}} = 3 \times \frac{v_{1} v_{2} v_{3}}{v_{3} v_{2} + v_{1} v_{3} + v_{1} {v_{2}}_{}} = \frac{3 v_{1} v_{2} v_{3}}{v_{3} v_{2} + v_{1} v_{3} + v_{1} v_{2}}$
Hence,
average speed = $\frac{3 v_{1} v_{2} v_{3}}{v_{3} v_{2} + v_{1} v_{3} + v_{1} {v_{2}}_{}}$

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