Question

In space, a 5.82×10^6kg rocket experiences a constant force of 6.94×10^9N in the direction of its movement. At time t=0s, it flies past a planet at a velocity of 57.2m/s. How far away is the rocket from the planet when t=25.4s?

Answer 1

The acceleration of the rocket can be found from the Newton's second law:

$$a = \dfrac{F}{m}$$

where $F = 6.94\times 10^9N$ and $m = 5.82\times 10^6kg$. Then the distance covered by the rocket is:

$$d = v_0t + \dfrac{at^2}{2} = v_0t + \dfrac{Ft^2}{2m}$$

where $v_0 = 57.2m/s$ and $t = 25.4s$. Thus, obtain:

$$d = 57.2\cdot 25.4 + \dfrac{6.94\times 10^9\cdot 25.4^2}{2\cdot5.82\times 10^6 } \approx 386110m$$

Answer. 386100 m.