Question

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?

Answer 1

(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.

Answer 2

(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.