If a body travels 2/5 Total distance with speed V1 and 3/ 5 of distance with speed V2 find average speed of the body?

Answer 1

time = distance / speed.

hence t1 = 2d/5 / v1 t2= 3d/5 / v2

now the average speed = distance / total time

= d/ ( d *(2/5v1 +3/5v2) )

= 1/ (2/5v1 + 3/5v2)

In other words the formula is almost the same as for electronics resistors in parallel = 1/( 1/R1 + 1/R2)

or optics f= 1/ (1/do + 1/di)

The only difference is that in this instance the terms are WEIGHTED.

ie it is a weighted geometric mean. Far from the only one you find in physics.

Answer 2

...

find total time

t = d/v

t total = 2/5V1 + 3/5V2 = (10V2 + 15V1) / 25V1V2 = (2V2 + 3V1) / 5V1V2

find avg speed

Vav = d / t = 1 / (2V2 + 3V1) / 5V1V2 = 5V1V2 / (2V2 + 3V1)

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Answer 3

pick a distance, say 100 (you could leave it as a variable, say D, but it cancels out in the end anyway. Easier to put in a number)

average speed is total distance divided by total time

then time to travel the first 40 meters (2/5 of total) is 40/V₁

time to travel the last 60 meters (3/5 of total) is 60/V₂

total time = (40/V₁) + (60/V₂) = (40V₂ + 60V₁) / V₁V₂

average speed = 100 / time = 100V₁V₂ / (40V₂ + 60V₁)

average speed = 5V₁V₂ / (2V₂ + 3V₁)

just to check, pick V₁ = 2 m/s and V₂ = 3 m/s

average speed = 5•6 /(6+6) = 30/12 = 2.5 m/s