# The surface area of a raindrop increases as it falls?

The surface area of a raindrop increases as it falls. Therefore, the air resistance to its falling increases as it falls until it reaches terminal velocity where the velocity becomes a constant. Assume that a raindrop has an initial velocity of -10 m/s and its acceleration is: [SEE IMAGE].

Define a piece-wise function that represents the velocity, in m/sec, of the raindrop at time t seconds. Relevance

You know that a(t) = dv/dt?

Why not plug that in?

Assuming v in m/s and t in s,

dv/dt = 0.9t - 9

dv = (0.9t - 9)dt

Either do the indefinite integral

v + C1 = 0.45t^2 - 9t + C2

v(t) = 0.45t^2 - 9t + C

And since v(0) = -10 m/s then C = -10

Or do the definite integral on left side from -10 to some v and from 0 to some t on the right side

v + 10 = 0.45t^2 - 9t + 0

Either way v(t) = 0.45t^2 - 9t - 10 from t = 0 to 10

v(t) = -55 for all t after 10.

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• Anonymous
8 months ago

In the first 10s a = dv/dt = 0.9t – 9

Integrate:

v = 0.45t² – 9t + C

Since v = -10 when t = 0

-10 = 0.45(0)² – 9(0) + C

C = -10

v = 0.45t² – 9t – 10

When t = 10s, v = 0.45(10)² – 9(10) – 10 = -55m/s

After 10s, v remains constant because a=0.

So the piece wise function (omitting units) for v is:

v(t) = 0.45t² – 9t – 10 when 0≤t≤10

. . . = 55 when 10<t

(Note.  As others have pointed out, the physics in the question is wrong, but this is simply a maths exercise ignoring the laws of physics!)

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• I don't believe the premise of the question.  At zero velocity it is a sphere.  As it gains speed it becomes more streamlined in shape.  The total surface area does go up but the FRONTAL surface area, which intersects the air, goes down.  The terminal velocity of a sphere is LOWER than the terminal velocity of a raindrop shape of equal volume and mass.

Now as V(t) = integral ( a(t) dt )  = 1/2 * 0.9 t^2 - 9t +c  ( 0<t<10)

by substituting t=0 you can observe the c must represent the initial velocity = -10 m/s^2  But that would give you a terminal velocity of - 55 m/s

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