Answer:
a)
, b)
![a = 96.667\,(m)/(s^(2))](https://img.qammunity.org/2021/formulas/engineering/college/q1bcano8kaphkn37kxjnx71yio9w0kpsxz.png)
Step-by-step explanation:
a) The maximum speed of the rocket is given by the Tsiolkovski's Equation:
![v =v_(o) - v_(ext)\cdot \ln \left((m)/(m_(o)) \right)](https://img.qammunity.org/2021/formulas/engineering/college/sal7m2xon2h36wq7dffzm10wrddtyq2yi4.png)
![v = 0\,(m)/(s) - (2900\,(m)/(s) )\cdot \ln \left((450\,kg)/(600\,kg) \right)](https://img.qammunity.org/2021/formulas/engineering/college/352aw9kcz5bvvsvhj1c99hrk8gjhazyy3l.png)
![v \approx 834.278\,(m)/(s)](https://img.qammunity.org/2021/formulas/engineering/college/2iiebb1iqg0xco272klygw2r4wugz48f14.png)
b) The acceleration is obtained by deriving the Tsiolkolski's Equation:
![a = -v_(ext)\cdot \left((1)/(m)\left) \cdot \dot m](https://img.qammunity.org/2021/formulas/engineering/college/6zrgz4k4w8mukayg1exues41hv2jiv4b0p.png)
The maximum acceleration occured when fuel is entirely consumed. Then:
![a = - \left(2900\,(m)/(s) \right)\cdot \left((1)/(450\,kg) \right)\cdot \left(-15\,(kg)/(s) \right)](https://img.qammunity.org/2021/formulas/engineering/college/jpfc7tsuutklpivm0lpyz89ftmjabnfqek.png)
![a = 96.667\,(m)/(s^(2))](https://img.qammunity.org/2021/formulas/engineering/college/q1bcano8kaphkn37kxjnx71yio9w0kpsxz.png)