A bullet is fired from a gun. The force on the bullet is given by $$F = 600 - 2 \times {10^5}t$$     where, $$F$$ is in newton and $$t$$ in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

Solution :
$${\text{Given}}\,F = 600 - \left( {2 \times {{10}^5}t} \right)$$
The force is zero at time $$t,$$ given by $$0 = 600 - 2 \times {10^5}t$$
$$\eqalign{ & \Rightarrow t = \frac{{600}}{{2 \times {{10}^5}}} = 3 \times {10^{ - 3}}\,{\text{seconds}} \cr & \therefore {\text{Impulse}} = \int\limits_0^t {Fdt} = \int\limits_0^{3 \times {{10}^{ - 3}}} {\left( {600 - 2 \times {{10}^5}t} \right)} dt \cr & = \left[ {600t - \frac{{2 \times {{10}^5}{t^2}}}{2}} \right]_0^{3 \times {{10}^{ - 3}}} \cr & = 600 \times 3 \times {10^{ - 3}} - {10^5}{\left( {3 \times {{10}^{ - 3}}} \right)^2} \cr & = 1.8 - 0.9 = 0.9\,Ns \cr} $$