Does anyone know the answer????

Question

The terminal velocity of a 3×10^−5 kg raindrop is about 9 m/s . Assume a drag force FD=−bv.

a) Determine the value of the constant b. Express your answer using one significant figure.

b= _____ kg/s

Determine the time required for such a drop, starting from rest, to reach 63% of terminal velocity. Express your answer using one significant figure.

_____________s

Answer 1

Let

positive direction be downwards and negative direction upwards,

m = 3×10^-5 kg be the mass of the raindrop,

v(t) be its velocity at time t,

v₀ = 9 m/s be its terminal velocity,

g = 9.81 m/s² be gravitational acceleration.

The raindrop experiences a downward (i.e. positive) gravitational force mg,

and upward (i.e. negative) drag force -bv.

(I assume that the minus sign was placed there so that b > 0.)

The total force at a time t is

F(t) = mg - bv(t)

a)

Terminal velocity is achieved then the total force is 0, i.e.,

0 = mg - bv₀

Therefore

b = mg/v₀

b)

Apply Newton's Second Law

F = m dv/dt

After substituting for F we get the differential equation

mg - bv = m dv/dt (1)

I do not know whether this is an exercise in physics or in ODE solving,

but let me just tell you one way of solving it from physical considerations.

Let t = 0 be the time the raindrop starts falling. Then

v(0) = 0

v(∞) = v₀

The equation (1) has v on the left and dv/dt on the right,

for which an exponential would work.

So we can guess that v(t) will follow exponential decay, and be of the

general form

v(t) = p + qe^rt for some constants p, q, r.

To satisfy v(0) = 0, we must have q = -p.

To satisfy v(∞) = v₀ , we must have r < 0 and p = v₀ .

So

v(t) = v₀ (1 - e^rt)

And from (a) we know that v₀ = mg/b.

So

v(t) = mg/b(1 - e^rt) (2)

Now we solve for r by plugging the general solution (2) into the equation (1)

mg - mg(1 - e^rt) = -m²gr/b e^rt

r = -b/m

So the solution to (1) is

v(t) = mg/b(1 - e^-bt/m) = v₀ (1 - e^-bt/m) (3)

To find the time t when v(t) = 63v₀ /100, we use the solution (3)

v₀ (1 - e^-bt/m) = 63v₀ /100

1 - e^-bt/m = 63/100

e^-bt/m = 37/100

t = -m/b ln(37/100) = -v₀ /g ln(37/100)

Substituting actual numbers

t = -9/9.81 ln(37/100) = 0.91 s