Question

"At time t = 0 a 2330-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 = 1.27 s .

What impulse does the engine exert on the rocket during the 1.50- s interval starting 2.00 s after the engine is fired?

Answer 1

`Imp = 5626.488\,(kg\cdot m)/(s)`

Step-by-step explanation:

First, it is required to model the function that models the increasing force in the +x direction:

`a =(781..25\,N)/((1.27\,s)^(2))`

`a = 484 (N)/(s^(2))`

The equation is:

`F_(x) = 484\,(N)/(s^(2))\cdot t^(2)`

The impulse done by the engine is given by the following integral:

`Imp=484\,(N)/(s^(2)) \int\limits^(3.50\,s)_(2\,s) {t^(2)} \, dt`

`Imp = 161.333\,(N)/(s^(2))\cdot [(3.50\,s)^(3)-(2\,s)^(3)]`

`Imp = 5626.488\,(kg\cdot m)/(s)`