Boat Drag Question: Acceleration After Rain Starts

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The discussion centers on calculating the acceleration of a boat with a mass of 250kg coasting at 3.00 m/s when it begins to accumulate rain at a rate of 10.0 kg/hr. The drag force acting on the boat is proportional to the square of its speed, given by F=0.5v^2. Participants are exploring how to incorporate the effects of rain accumulation into the momentum equations, noting that momentum is not conserved due to the added mass. A key point is the calculation of the force from rain accumulation, which affects the boat's horizontal momentum. The final acceleration can be determined by combining the drag force and the force from the rain accumulation, leading to a negative acceleration as the boat slows down.
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A boat of mass 250kg is coasting, with its engine in neutral, through the water at speed 3.00m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0kg/hr. Assume that the boat is subject to a drag force due to water resistance. The drag is proportional to the square of the speed of the boat, in the form F=0.5v^2. What is the acceleration of the boat just after the rain starts?

I've been approaching this question from a few angles and don't seem to be getting anywhere. I found the answer to the 1st part using conservation of momentum, but since momentum is no longer conserved I don't know what to do. I'd really appreciate some help with this.
 
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a = \frac{0.5v^2}{m}

But I don't see how you could use the other problem data.
 
vpea said:
A boat of mass 250kg is coasting, with its engine in neutral, through the water at speed 3.00m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0kg/hr. Assume that the boat is subject to a drag force due to water resistance. The drag is proportional to the square of the speed of the boat, in the form F=0.5v^2. What is the acceleration of the boat just after the rain starts?

I've been approaching this question from a few angles and don't seem to be getting anywhere. I found the answer to the 1st part using conservation of momentum, but since momentum is no longer conserved I don't know what to do. I'd really appreciate some help with this.
SOLUTION HINTS:
{Rain Accumulation Rate} = (10.0 kg/hr) = (2.7778e(-3) kg/sec)
{Horizontal Boat Speed} = (3.00 m/s)

From Conservation of Momentum between Boat and Rain Drops:
{Change in Boat's Horizontal Momentum DUE TO RAIN ACCUMULATION} =
= -{Change in Rain's Horizontal Momentum when Accumulating in Boat}

::: ⇒ -{FORCE on Boat from Rain Accumulation} = Frain =
= -{RATE OF CHANGE in Boat's Horizontal Momentum DUE TO RAIN ACCUMULATION} =
= {RATE OF CHANGE in Rain's Horizontal Momentum when Accumulating in Boat} =
= {Rain Accumulation Rate}*{Horizontal Boat Speed} =
= (2.7778e(-3) kg/sec)*(3.00 m/s) =
= (8.3334e(-3) N)

{Acceleration of Boat} = {Net Force on Boat}/{Mass of Boat} =
= {-(Drag Force) - (Frain)}/{Mass of Boat}
= {-0.5v^2 - (Frain)}/{Mass of Boat} =
= { (-0.5)*(3.00 m/s)^2 - (Frain) }/{Mass of Boat} =
Determine boat acceleration (negative because boat is slowing) from this last equation using values given and/or computed previously.


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