A particle moving in a straight line covers half the distance with speed of 3m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is?

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 All the best!!!!!

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

 

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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.

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.🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏

Hello Student

Let the total distance covered be 2x.

Let t₁ be the time taken the first half of the distance i.e. x.

Let the other half of the distance (x) be covered in two equal intervals of time t₂.

x = 4.5t₂ + 7.5t₂
x = 12t₂
t₂ = x/12

Total Time (T) = t₁+ 2t₂= x/3 + x/6 = x/2

Average Speed = total distance travelled / total time taken
Average speed = (2x)/x/2 = 4ms^-1.

I hope this answer will help you.