Answer
The impulse on the ball due to the kick is $~~9.0~kg~m/s$
Work Step by Step
We can find the impulse on the ball:
$J = \int^{3.0\times 10^{-3}}_{0}[(6.0\times 10^6)t-(2.0\times 10^9)t^2)]~dt$
$J = [(3.0\times 10^6)t^2-(\frac{2.0}{3}\times 10^9)t^3)]\Big\vert^{3.0\times 10^{-3}}_{0}$
$J = [(3.0\times 10^6)(3.0\times 10^{-3})^2-(\frac{2.0}{3}\times 10^9)(3.0\times 10^{-3})^3]-[(3.0\times 10^6)(0)^2-(\frac{2.0}{3}\times 10^9)(0)^3)]$
$J = (27.0-18.0)-(0)$
$J = 9.0~kg~m/s$
The impulse on the ball due to the kick is $~~9.0~kg~m/s$.
