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.

Attachment image

4 Answers

Relevance
  • 3 weeks ago
    Best answer

    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.

  • Anonymous
    3 weeks 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!)

  • 3 weeks ago

    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

  • Henry
    Lv 5
    3 weeks ago

    No it doesn't unless it is tested. And this sounds like homework.

Still have questions? Get answers by asking now.