Answer:
[tex]v(10\,s) \approx 775.387\,\frac{m}{s}[/tex]
[tex]v(20\,s)\approx 1905.350\,\frac{m}{s}[/tex]
[tex]v(30\,s) \approx 4023.595\,\frac{m}{s}[/tex]
Explanation:
The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:
[tex]v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}[/tex]
Where:
[tex]v_{o}[/tex] - Initial speed of the rocket, in m/s.
[tex]v_{ex}[/tex] - Exhaust gas speed, in m/s.
[tex]m_{o}[/tex] - Initial total mass of the rocket, in kg.
[tex]m[/tex] - Current total mass of the rocket, in kg.
Let assume that fuel is burned linearly. So that,
[tex]m(t) = m_{o} + r\cdot t[/tex]
The initial total mass of the rocket is:
[tex]m_{o} = 750\,kg[/tex]
The fuel consumption rate is:
[tex]r = -\frac{600\,kg}{30\,s}[/tex]
[tex]r = -20\,\frac{kg}{s}[/tex]
The function for the current total mass of the rocket is:
[tex]m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t[/tex]
The speed function of the rocket is:
[tex]v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}[/tex]
The speed of the rocket at given instants are:
[tex]v(10\,s) \approx 775.387\,\frac{m}{s}[/tex]
[tex]v(20\,s)\approx 1905.350\,\frac{m}{s}[/tex]
[tex]v(30\,s) \approx 4023.595\,\frac{m}{s}[/tex]
