Boat Drag Question: Acceleration After Rain Starts

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SUMMARY

The discussion centers on calculating the acceleration of a boat with a mass of 250 kg, coasting at a speed of 3.00 m/s, when rain begins to accumulate at a rate of 10.0 kg/hr. The drag force acting on the boat is defined by the equation F = 0.5v². The net force on the boat is determined by the sum of the drag force and the force due to rain accumulation, leading to the formula for acceleration: a = {-(0.5 * (3.00 m/s)²) - (8.3334e(-3) N)}/{250 kg}. This results in a negative acceleration, indicating that the boat is slowing down as the rain accumulates.

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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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[tex]a = \frac{0.5v^2}{m}[/tex]

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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