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# Integration

When a raindrop falls in still air, its acceleration is given by g-0.5v, where g is ate acceleration due to gravity  and v is the velocity. We then have a = dv/dt = g-0.5v.

a) Taking g as 10ms^-2, determine the velocity of the raindrop after two seconds if it starts falling from rest.

Ans: 12.64ms^-1

b) When the acceleration is less than 0.01mm s^-2, the raindrop will effectively be travelling at uniform velocity.
i) At what velocity does this situation arise?

Ans: 19.98ms/s

ii) How long does it take to reach this state?

Ans: 27.63s

Just need to know the working out, thanks.

### 1 Answer by Expert Tutors

Arturo O. | Experienced Physics Teacher for Physics TutoringExperienced Physics Teacher for Physics ...
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I will work part (a), but for part (b) you need to fix a few numbers.  The acceleration cannot be 0.01 mm/s.  Did you mean 0.01 mm/s2, or perhaps even 0.01 m/s2 ?  Fix that.  Also, the answer to (b)(i) is a speed, and therefore cannot have units of ms/s.  Fix that too.  Double check the number and the units.

Anyway, here goes with part (a):

(a)

dv/dt = 10 - 0.5v

dv/(10 - 0.5v) = dt

-0.5 dv/(10 - 0.5v) = -0.5 dt

Integrate both sides.

ln(10 - 0.5v) = -0.5t + c

Take e() of both sides.

10 - 0.5v = ke-0.5t

v(t) = 2(10 - ke-0.5t)

v(0) = 0  [starts from rest]

0 = 2(10 - k) ⇒ k = 10

v(t) = 20(1 - e-0.5t)

v(2) = 20[1 - e-0.5(2)] ≅ 12.64 m/s

Fix the things I mentioned above and I will be happy to work part (b).