Let total distance=d
   Total time=T
Then in  d/2.......
t1=d/6

t2=d/(4*4.5)  +  d/(4*7.5)
=4d/45

T=t1+t2=d/6+4d/45 =23d/90

Average speed =t. Dis/t. Time= d/T =3.91
m/s........ANS
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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
 Ttotal = 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
 
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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.
						

							
all the answers are waste of time 
there is a formulae 2v1*v2/v1+v2 if the particle travels in 2 equal time intervals so lets take second distance 4.5+7.5/2 =12m/s so 2*v1*v2/v1+v2 = 2*3*6/6+3 =36/9=4.0 m/s 
						

							
all the answers are waste of time 
there is a formulae 2v1*v2/v1+v2 if the particle travels in 2 equal time intervals so lets take second distance 4.5+7.5/2 =12m/s so 2*v1*v2/v1+v2 = 2*3*6/6+3 =36/9=4.0 m/s 
answered by a 8th grade
						

							Avg.speed =total distance /total time taken Avg. Speed in 2nd half=(sT +ST)/2T                               = (4.5T+7.5T)/2T=6m/sTotal avg.Speed =  D/(d/2)/3+(d/2)/6                        =12D/3D=4m/s
						

							V1=3m/sV2=4.5 m/sV3=7.5 m/sTherefore formula for dist covered in unequal speed in equal intervals of time=2V1(V1+V3)/V1+V2+V3=2x3(4.5+7.5)/3+4.5+7.5=4.875
						

							Average Speed (half distance formulae):=>2×[V1×V2] / V1+V2.=>[3×(4.5+7.5)] / 6+3=>[3×12]/9=>36/9=>4.0 m/s.🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏