Respuesta :
Explanation:
Given that,
Mass of the rocket, m = 2150 kg
At time t=0 a rocket in outer space fires an engine that exerts an increasing force on it in the +x−direction. The force is given by equation :
[tex]F=At^2[/tex]
Here F = 888.93 N when t = 1.25 s
(c) We can find the value of A first as :
[tex]F=At^2\\\\A=\dfrac{F}{t^2}\\\\A=\dfrac{888.93}{(1.25)^2}\\\\A=568.91\ N/s^2[/tex]
The value of A is [tex]568.91\ N/s^2[/tex].
(a) Let J is the impulse does the engine exert on the rocket during the 4.0 s interval starting 2.00 s after the engine is fired. It is given in terms of force as :
[tex]J=\int\limits {F{\cdot} dt}[/tex]
Limits will be from 2 s to 2+ 4 = 6 s
It implies :
[tex]J=\int\limits^6_2 {At^2{\cdot} dt}\\\\J=A\int\limits^6_2 {t^2{\cdot} dt}\\\\J=A\dfrac{t^3}{3}|_2^6\\\\J=568.91\times \dfrac{1}{3}\times (6^3-2^3)\\\\J=39444.42\ Ns[/tex]
(b) Impulse is also equal to the change in momentum as :
[tex]J=m\Delta v\\\\\Delta v=\dfrac{J}{m}\\\\\Delta v=\dfrac{39444.42}{2150}\\\\\Delta v=18.34\ m/s[/tex]
Hence, this is the required solution.
