Drag force acting on a boat

I'm doing this question:

A boat of mass 250kg is coasting, with its engine in neutral, through the water at speed 1.00m/s when it starts to rain with incredible intensity. The rain is falling vertically, and it accumulates in the boat at the rate of 100kg/hr.
The drag is proportional to the square of the speed of the boat, in the form Fd=bv2 where b= 0.5 N⋅ s2/m2. What is the acceleration of the boat just after the rain starts? Take the positive x axis along the direction of motion.

I've differentiated momentum wrt time so I could find the net force to get $\frac{dp}{dt}= kv + (m_o + kt)\frac{dv}{dt}$ where k is the rate at which the boat is filling with water, v is the velocity of the boat, $m_0$ is the initial mass and t is time. I'm not really sure where to go from here. Any help?

If this is A level you won't be expected to solve a differential equation...

you know that while it is coasting it is at constant velocity so you can work out the total drag acting on it under 'normal' conditions, and hence find the force acting to propel the boat. When it starts to fill up with water, why does the boat accelerate down in the first place?

It's not A-level, it's undergraduate.

What do you mean by 'accelerate down'?

woops, I originally wrote slow down, then changed my mind and wrote accelerate.... apparently I didn't edit it properly.

If it's undergrad then I can't help you probably. The way I was thinking about it doesn't work when I actually tried it . I'm not sure I understand the question very well actually, as surely this just means the mass increases and none of the forces acting on the boat are actually related to the mass? I.e. drag force depends only on current speed and I assume thrust is kept constant...

Anyway, sorry I can't help you.

I'm doing this question:

A boat of mass 250kg is coasting, with its engine in neutral, through the water at speed 1.00m/s when it starts to rain with incredible intensity. The rain is falling vertically, and it accumulates in the boat at the rate of 100kg/hr.
The drag is proportional to the square of the speed of the boat, in the form Fd=bv2 where b= 0.5 N⋅ s2/m2. What is the acceleration of the boat just after the rain starts? Take the positive x axis along the direction of motion.

Do you have the answer? I get $a=-1.88 \times 10^{-3}$ m/s^2 via:

1. $F=\frac{dp}{dt}=\frac{d(mv)}{dt}$
2. Knowing that the only external force on the boat is the drag
3. Rearranging for $a$ and plugging the numbers in

Do you have the answer? I get $a=-1.88 \times 10^{-3}$ m/s^2 via:

1. $F=\frac{dp}{dt}=\frac{d(mv)}{dt}$
2. Knowing that the only external force on the boat is the drag
3. Rearranging for $a$ and plugging the numbers in

Ahh I finally got it! $-2.11 \times 10^{-3} ms^{-2}$

Our initial answer was using positive bv, it needed to be negative because of the drag opposing the motion of the boat.

Right. In fact, I should have got that value initially (I had taken account of the direction of drag) but I simply neglected the -ve sign when plugging my numbers in to the calculator. Sorry.