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?
- Andrew SmithLv 73 years ago
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.
- Old Science GuyLv 73 years ago
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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- billrussell42Lv 73 years ago
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