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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. 
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1 Answer

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):
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).