A spacecraft engine creates \(53.2 \mathrm{MN}\) of thrust with a propellant velocity of \(4.78 \mathrm{~km} / \mathrm{s}\). a) Find the rate \((d m / d t)\) at which the propellant is expelled. b) If the initial mass is \(2.12 \cdot 10^{6} \mathrm{~kg}\) and the final mass is \(7.04 \cdot 10^{4} \mathrm{~kg},\) find the final speed of the spacecraft (assume the initial speed is zero and any gravitational fields are small enough to be ignored). c) Find the average acceleration till burnout (the time at which the propellant is used up; assume the mass flow rate is constant until that time).

We are given the thrust force \(F = 53.2 \times 10^{6}\,\text{N}\) and the propellant velocity \(v_{propellant} = 4.78 \times 10^{3}\,\text{m/s}\). To calculate the mass flow rate \(\frac{dm}{dt}\), we'll make use of the second law relationship: $$F = \frac{d(mv)}{dt}$$ Here, \(mv\) represents the change in momentum of the rocket. Since the propellant velocity is constant, we can write this equation as: $$F = v_{propellant} \times \frac{dm}{dt}$$ Solving for the mass flow rate \(\frac{dm}{dt}\), we have: $$\frac{dm}{dt} = \frac{F}{v_{propellant}} = \frac{53.2 \times 10^{6}\,\text{N}}{4.78 \times 10^{3}\,\text{m/s}} \approx 11.13\times 10^{3} \,\text{kg/s}$$

Now we'll make use of the conservation of momentum and the Rocket Propulsion Equation to calculate the final speed of the spacecraft (\(v_f\)). The equation is given by: $$v_{f} = v_{propellant} \ln \left(\frac{m_{initial}}{m_{final}}\right)$$ Plugging in the values given, we obtain: $$v_{f} = 4780\, \text{m/s} \times \ln \left(\frac{2.12 \times 10^6\, \text{kg}}{7.04 \times 10^4\, \text{kg}}\right) \approx 17,306\, \text{m/s}$$ So the final speed of the spacecraft is \(17,306\, \text{m/s}\).

Now we'll calculate the average acceleration of the spacecraft until burnout, assuming a constant mass flow rate. Since the mass flow rate is constant, we can use the following equation to calculate the burnout time \(t_{burnout}\): $$t_{burnout} = \frac{\Delta m}{\frac{dm}{dt}} = \frac{2.12 \times 10^6\,\text{kg} - 7.04 \times 10^4\,\text{kg}}{11.13\times 10^{3} \,\text{kg/s}} \approx 178.35\, \text{s}$$ To calculate the average acceleration (\(a_{avg}\)), we'll consider the initial and final velocities (\(v_{i} = 0\,\text{m/s}\) and \(v_{f} = 17,306\, \text{m/s}\)) and the burnout time, and then use the equation: $$a_{avg} = \frac{v_{f} - v_{i}}{t_{burnout}} = \frac{17,306\,\text{m/s}}{178.35\,\text{s}} \approx 97.03\, \text{m/s}^{2}$$ So the average acceleration till burnout is about \(97.03\, \text{m/s}^{2}\).