Integration

Question

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. 

Answer 1

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).