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

Answer:

$I=4090.8Ns$

Explanation:

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

$A=\frac{F}{t^2}=\frac{781.25N}{(1.49s)^2}=351.9N/s^2$

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

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

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

$I = (351.9N/s^2) (\frac{(3.5s)^3}{3}-\frac{(2s)^3}{3})=4090.8Kgm/s$