What is the speed of the boat after time 2.00 hr has passed?

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The discussion focuses on calculating the speed of a boat after 2 hours while it accumulates rainwater. Initially, the boat has a mass of 250 kg and travels at 3.00 m/s. After 2 hours, with rain accumulating at a rate of 10.0 kg/hr, the total mass increases to 270 kg. The drag force acting on the boat is defined as F_d = 0.5 v^2, impacting its acceleration. The discussion also addresses the conservation of momentum in the presence of drag forces.

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Can you please help me out with the second part of this question? I got the first two parts. Its PART C that I am not getting. Thanks!

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.

Part A
What is the speed of the boat after time 2.00 hr has passed? Assume that the water resistance is negligible.

Part B
Now assume that the boat is subject to a drag force F_d due to water resistance. Is the component of the total momentum of the system parallel to the direction of motion still conserved?

Part C
The drag is proportional to the square of the speed of the boat, in the form F_{\rm d}= 0.5 v^2. What is the acceleration of the boat just after the rain starts? Take the positive x-axis along the direction of motion.
Express your answer in meters per second per second.


Its Part C that I am not getting. Please help. Thanks!


Thank-you very much for all your time and effort!
 
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Use Newton's 2nd law.
 


Is the component of the total momentum of the system parallel to the direction of motion still conserved?

I do not get part B, the part where the system is parallel to the direction of motion. can anyone explain??

do i think of the boat and the drag force as a collision? therefore, total momentum is conserved?
 

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