RelientKayers asked:
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 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.
The drag is proportional to the square of the speed of the boat, in the form F_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.
Make a momentum balance in the direction of motion (the x-direction).
The change of the momentum equals the forces acting on the boat. There is only the drag force, which acts against the direction of motion. :
d(m·v)/dt = -F_d = – k·v²
where k = 0.5 Ns²/m²
Expand the time derivative on the LHS, by applying product rule of differentiation:
v·dm/dt + m·dv/dt = – k·v²
where dm/dt = 10kg/hr is the change of the mass of the boat
The acceleration is the time derivative of time velocity:
a = dv/dt = – (k·v² + v·dm/dt) / m
For the moment,w hen the rain starts all the value on the right hand side are known, thus:
a₀ = – (k·v₀² + v₀·dm/dt) / m₀
= – (0.5Ns²/m² · (3m/s)² + 3m/s · 10kg/3600s ) / 250kg = 18.03·10⁻³ m/s²
▬ thus before the rain her engine power P is constant; and elementary work done du=P*dt is work done against drag force F_d = 0.5v^2, so du=F_d*dx;
P*dt=0.5v^2*dx; or; P=0.5*v^3;
♠ let t=0 when the rain starts; her initial speed v(0)=3m/s; we also assume that her engine power P is constant; thus for t0 elementary balance of energies looks:
du = dE +dW, where du = P*dt is work of engine in time dt, dE = d(0.5*m*v^2) is change in kinetic energy, dW =F_d*(v*dt) is elementary work of drag force, v*dt is elementary path in time dt; thus
P*dt = 0.5*dm*v^2 +m*v*dv +0.5*v^2 *v*dt; or;
♣ P = 0.5*m’*v^2 +m*v*v’ +0.5*v^3, where v’ is acceleration of the boat, m’=10 kg/hr =10/3600 kg/s is given rate of mass increase of the boat;
♦ for t=0 we know that P=0.5*v^3, so we get:
0 = 0.5m’*v^2 +m*v*v’, hence v’= -0.5v*(m’/m) =
= -0.5*3*(10/3600)/250 = -(5/3)·10^(-5) m/s^2;