# Boat Drag question

1. Apr 11, 2005

### vpea

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.

2. Apr 11, 2005

### ramollari

$$a = \frac{0.5v^2}{m}$$

But I don't see how you could use the other problem data.

3. Apr 11, 2005

### xanthym

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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Last edited: Apr 11, 2005