Question

a particle travells first half of the total distance with the speed v1. In second half distance constant speed in 1/3rd time is v2 and in remaining 2/3rd time constant speed is v3. Find the average speed durind the complete journey​

Answer 1

Answer:

| Vavg | = 2v₁ ( v₂  + 2V₃ ) / ( v₂  + 2V₃ + 3v₁  )

Explanation:

From AC, v₁ = t/d

Thus t = d/v₁      -------  ( 1 )

From CD,  v₂ = CD/t/3

Thus v₂ = 3CD/t    -------- ( 2 )

From DB, V₃ = DB/2t/3   -------- ( 3 )

Now,

CD + DB = d

v₂t / 3) +  2V₃t / 3 = d

t/3 [ v₂  + 2V₃ ] = d

Thus,  t = 3d / v₂  + 2V₃

Now, Average speed  =  total distance / total time

Vavg = ( d + d ) / tAC + tCD + tDB

substituting value of t from Eq.(i), we have

Vavg = 2d / d / v₁ + 3d / v₂  + 2V₃

Solve a little to find the final answer

| Vavg | = 2v₁ ( v₂  + 2V₃ ) / ( v₂  + 2V₃ + 3v₁  )

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