let total distance of path is S ...

for first half , time taken if t

speed = distance/time

3 = S/2t

t = S/6      .............1

for second half , time taken is T ...

for first  T/2 it moves with 4.5

speed = distance/time

4.5 = 2x/T

 x = 4.5T/2   ...........2

for second T/2 it moves with 7.5

7.5 = 2y/T

y = 7.5T/2       ..........3

we have , x+y = S/2so

(7.5+4.5)T = S/2

T = S/24        ...............4

total time taken during its motion is T+t

 T_total = T+t = S/24+S/6 = 5S/24

total distance = S

average speed = total distance/total time = S/5S/24 = 24/5 = 4.8m/s

approve if u like my ans

hi

let the distanc covered with speed 3m/s be x.mts

time taken is x/3sec.

ii)second half(x mts)

time taken to travel second half will be 18x/11.25 sec

avg. speed=total distanc /totaltime

avg.speed=1.9m/s

plzz. prove!!!!!!!!!!!!!!!!!!!!!!!!!!

Let total distance be S. Total time be T and V1= 3m/s be the velocity for which it covers the first half of the distance and V2 = 4.5 m/s and V3 = 7.5 m/s be the velocities in the second half of the distance. Let <v> be the average velocity that has to be calculated.

Let the other half be completed in times t1 and t2 (t1= t2=t, from the question).

For Ist half of the distance:

S/2= V1(T-2t) 

=> T = S/2V1 + 2t              ....(1)

For 2nd half of the distance:

S/2 = (V2 + V3)2t

=> S = 4(V2 + V3)t             ....(2)

From (1) and (2).

T = {2(V2 + V3)/V1    + 2}t      ....(3)

we know that,

<v> = total distance/total time

so,

<v> = S/T

Putting the values of  T and S from (3) and (2),

<v> = 4(V2 + V2)t /{2(V2 + V3)/V1   +2 }t

=> <v> = 2V1 (V2 + V3)/(2V1 + V2 + V3)

putting the values of V1, V2 and V3.

<v> = 4 m/s.