5.A particle travels half the distance with a velocity of 6 m/s. The r

1.	5.A particle travels half the distance with a velocity of 6 m/s. The remaining half distance iscovered with a velocity of 4 m/s for half the time and with a velocity 8 m/s for the rest of thehalf time. What is the velocity of the particle averaged over the whole time of the motion?(A) 9 m/s) 6 m/s(C) 5.35 m/s(D) 5 m/s.
Answer» Let the total distance be D. the point transversed half the distance by velocity 6m/s. Then the time required by it to do so = (D/2)/6=D/12 The remaining half was transversed by travelling at 4m/sfor halftime and 6m/svelocity for another half time. Let the time to cover the second half of the journey = t Then4t/2+8t/2= D/2(Second Half of journey) So t=D/(4+8)=D/12 The total time required for the journey is (D/2(6))+(D/(4+8) The Average or mean Velocity = D/((D/2(6)) + (D/(4+8 ) =1/((1/2(6)) + (1/(4+8 ) ==2(6)(4+8)/(12+4+8)=12(12)/(24)=6m/s (for your reference)Let the total distance be D. Let the point transversed half the distance by velocity V1. Then the time required by it to do so = (D/2)/V1=D/2V1 The remaining half was transversed by travelling at V2for halftime and V3velocity for another half time. Let the time to cover the second half of the journey = t ThenV2t/2+V3t/2= D/2(Second Half of journey) So t=D/(V2+V3) The total time required for the journey is (D/2V1)+(D/(V2+V3)) The Average or mean Velocity = D/((D/2V1) + (D/(V2+V3)) ) =1/((1/2V1) + (1/(V2+V3)) ) =2V1(V2+V3)/(2V1+V2+V3)----------(ans)

5.A particle travels half the distance with a velocity of 6 m/s. The remaining half distance iscovered with a velocity of 4 m/s for half the time and with a velocity 8 m/s for the rest of thehalf time. What is the velocity of the particle averaged over the whole time of the motion?(A) 9 m/s) 6 m/s(C) 5.35 m/s(D) 5 m/s.

Answer»

Let the total distance be D.

the point transversed half the distance by velocity 6m/s.

Then the time required by it to do so = (D/2)/6=D/12

The remaining half was transversed by travelling at 4m/sfor halftime and 6m/svelocity for another half time.

Let the time to cover the second half of the journey = t

Then4t/2+8t/2= D/2(Second Half of journey)

So t=D/(4+8)=D/12

The total time required for the journey is (D/2(6))+(D/(4+8)

The Average or mean Velocity = D/((D/2(6)) + (D/(4+8 ) =1/((1/2(6)) + (1/(4+8 ) ==2(6)(4+8)/(12+4+8)=12(12)/(24)=6m/s

(for your reference)Let the total distance be D.

Let the point transversed half the distance by velocity V1.

Then the time required by it to do so = (D/2)/V1=D/2V1

The remaining half was transversed by travelling at V2for halftime and V3velocity for another half time.

Let the time to cover the second half of the journey = t

ThenV2t/2+V3t/2= D/2(Second Half of journey)

So t=D/(V2+V3)

The total time required for the journey is (D/2V1)+(D/(V2+V3))

The Average or mean Velocity = D/((D/2V1) + (D/(V2+V3)) ) =1/((1/2V1) + (1/(V2+V3)) ) =2V1(V2+V3)/(2V1+V2+V3)----------(ans)