Posted by **Zac** on Thursday, July 26, 2012 at 11:59am.

Starting from rest, a 6.70 kg object falls through some liquid and experiences a resistive (drag)

force that is linearly proportional to the velocity of the object. It's measured that the object reaches

half its terminal speed at 4.50 s. a) What is the terminal speed for this object in this liquid? b) At

what time will the speed be 3/4 of the terminal speed? c) How far will the object have traveled in the

first 4.50 s of falling?

- Physics -
**Elena**, Thursday, July 26, 2012 at 2:52pm
(a)

In fluid dynamics, an object is moving at its terminal velocity if its speed is constant due to the restraining force exerted by the fluid through which it is moving. Therefore,

m•g=r•V, where r is drag coefficient,and V is the terminal force.

V= m•g/r.

Now for accelerated motion,

m•a=m•g-r•v,

dv/dt =g –r•v/m,

Δv/Δt= g –r•v/m,

Since Δv =v(fin)-v(initial), and

v(initial) = 0,

v(fin)=V/2 =>

Δv = V/2.

Then

V/2•Δt= g – r•V/2•m,

m•g/2•r• Δt = g - r•m•g/m•r•2

r=m/ Δt =6.7/4.5 = 1.49 kg/s.

V=mg/r= 67•9.8•4.5/6.7=44.1 m/s.

(b) v=3•V/4.

Δv/Δt= g –r•v/m,

3V/4•Δt =g - r•3•V/4•m,

Solve for Δt,

Δt = 3m/r=3•6.7/1.49 =13.49 s.

- Physics -
**Elena**, Thursday, July 26, 2012 at 3:06pm
(c)

m•a = m•g - m•v,

a = g-r•v/m,

Multiply this equation by Δt.

a• Δt =g• Δt - r•v• Δt /m,

Since a• Δt =Δv=V/2 =22.05 m/s, and

v• Δt = Δs, we obtain

Δs =(m/r)(g• Δt – Δv)=

=(6.7/1.49)(9.8•4.5-22.05)=99 m.

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