Answer:
[tex]I=4090.8Ns[/tex]
Explanation:
Since our equation is [tex]F=At^2[/tex] and F=781.25 N when t=1.49s, we have:
[tex]A=\frac{F}{t^2}=\frac{781.25N}{(1.49s)^2}=351.9N/s^2[/tex]
The impulse of a force is [tex]I=\int\limits {F} \, dt[/tex], so for our case we have:
[tex]I=\int\limits^{t_f}_{t_i} {At^2} \, dt=A (\frac{t_f^3}{3}-\frac{t_i^3}{3})[/tex]
For our case [tex]t_i=2s[/tex] and [tex]t_f=2s+1.5s=3.5s[/tex], so we have:
[tex] I = (351.9N/s^2) (\frac{(3.5s)^3}{3}-\frac{(2s)^3}{3})=4090.8Kgm/s[/tex]
