Answer:
a) [tex]v(0.5\,h) = 0.833\,\frac{m}{s}[/tex], b) Yes, c) [tex]a = -0.4\,\frac{m}{s^{2}}[/tex]
Explanation:
a) A boat travelling in neutral means that boat is not experimenting any force from engine, the equation of equilibrium is:
[tex]\dot m \cdot v + m \cdot a = 0[/tex]
[tex]\dot m \cdot v + (m_{o} + \dot m\cdot t) \cdot \frac{dv}{dt} = 0[/tex]
The following differential equation is constructed:
[tex]-\dot m \cdot v = (m_{o} + \dot m \cdot t)\cdot \frac{dv}{dt}[/tex]
[tex]-\frac{\dot m\,dt}{m_{o}+\dot m\cdot t} = \frac{dv}{v}[/tex]
[tex]-\int\limits^{t}_{0} {\frac{\dot m}{m_{o}+\dot m \cdot t} } \, dt = \ln \left |\frac{v}{v_{o}} \right|[/tex]
[tex]\ln \left |\frac{m_{o}}{m_{o}+\dot m \cdot t} \right| = \ln \left|\frac{v}{v_{o}} \right|[/tex]
[tex]v(t) = v_{o} \cdot \frac{m_{o}}{m_{o}+\dot m\cdot t}[/tex]
The speed at t = 0.5 hours is:
[tex]v(0.5\,h) = (1\,\frac{m}{s} )\cdot \left[\frac{250\,kg}{250\,kg+(100\,\frac{kg}{h} )\cdot (0.5\,h)} \right][/tex]
[tex]v(0.5\,h) = 0.833\,\frac{m}{s}[/tex]
b) Since there is no external force exerted on the boat, the component of the total momentum of the system parallel to the direction of motion is conserved.
c) The acceleration of the boat after the rain starts is:
[tex]a = -\frac{\dot m\cdot v}{m}[/tex]
[tex]a = - \frac{(100\,\frac{kg}{h} )\cdot (1\,\frac{m}{s} )}{250\,kg}[/tex]
[tex]a = -0.4\,\frac{m}{s^{2}}[/tex]
