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# 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?

10 years ago
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!!!!!
10 years ago

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

approve if u like my ans

10 years ago

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!!!!!!!!!!!!!!!!!!!!!!!!!!

10 years ago

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.

4 years ago
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
4 years ago
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
4 years ago
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
3 years ago
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
3 years ago
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.🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏 Yash Chourasiya
one year ago
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.