Question

At time t = 0 a 2230-kg rocket in outer space fires an engine that exerts an increasing force on it in the +x-direction. This force obeys the equation Fx =At2 , where t is time, and has a magnitude of 781.25 N when t = 1.49s. What impulse (in kg m/s) does the engine exert on the rocket during the 1.50-s interval starting 2.00 s after the engine is fired?

Answer 1

`I=4090.8Ns`

Step-by-step explanation:

Since our equation is
`F=At^2` and F=781.25 N when t=1.49s, we have:

`A=(F)/(t^2)=(781.25N)/((1.49s)^2)=351.9N/s^2`

The impulse of a force is
`I=\int\limits {F} \, dt`, so for our case we have:

`I=\int\limits^(t_f)_(t_i) {At^2} \, dt=A ((t_f^3)/(3)-(t_i^3)/(3))`

For our case
`t_i=2s` and
`t_f=2s+1.5s=3.5s`, so we have:

`I = (351.9N/s^2) (((3.5s)^3)/(3)-((2s)^3)/(3))=4090.8Kgm/s`