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.

The first part of the question is missing and it is:

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 3.00 m/s when it starts to rain. The rain is falling vertically, and it accumulates in the boat at the rate of 10.0 kg/hr.

a(o) = 18.03 x 10^(-3) m/s^(-2)

Explanation:

First of all, if we make a momentum balance in the direction of motion (the x-direction), we'll discover that the change of the momentum will be equal to the forces acting on the boat.

Hence, there is only the drag force, which acts against the direction of motion as:

d(m·v)/dt = - k·v² (since f_d = - k·v²)

where k = 0.5 Ns²/m²

Now let's simplify the time derivative on the Left, by applying product rule of differentiation and we obtain:

v·dm/dt + m·dv/dt = - k·v²

where dm/dt = 10kg/hr (this is the change of the mass of the boat)

Furthermore, acceleration is the time derivative of time velocity, Thus;

a = dv/dt = - (k·v² + v·dm/dt) / m

Now, for the moment,when the rain starts; since we know all the values on the right hand side, let's solve for the acceleration ;

a(o) = - (ko·vo² + vo·dm/dt) / mo

= -[ (0.5(3)²) + (3x10/3600)/250

(dt = 3600secs because 1hr = 3600 secs)

So a(o) = 18.03 x 10^(-3) m/s^(-2)

i think it is non rem sleep

it may be wrong but i think its a