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.
- RealProLv 78 months agoFavourite 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.
- Anonymous8 months ago
In the first 10s a = dv/dt = 0.9t – 9
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!)
- Andrew SmithLv 78 months 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