A constant net force ${\vec F}$ is applied on a body during a time interval $\Delta t = {t_2}-{t_1}$ and the body undergoes a displacement in the direction of force. The work done is the product of the force and the displacement but the *impulse* is the product of the force and the time interval. Therefore impulse is the measure of how long the force is applied. The impulse $J$ is

\[\vec J = {\vec F}\Delta t \tag{1} \label{1}\]

Note that impulse is a vector quantity. Further investigation of Eq. \eqref{1} gives,

\[\begin{align*} \vec J &= {\vec F}\Delta t\\ &= m\vec a \Delta t\\ &= m\frac{{\Delta \vec v }}{{\Delta t}}\Delta t\\ &= m\Delta \vec v \\ &= m({\vec v _f} - {\vec v _i})\\ {\rm{or,}}\quad \vec J &= m{\vec v _f} - m{\vec v _i} \tag{2} \label{2} \end{align*}\]

Linear momentum is the product of mass and linear velocity of a body denoted by $p$. So $\vec p = m{\vec v}$ and in Eq. \eqref{2}, the initial linear momentum is ${\vec p _i}= m{\vec v_i}$ and final linear momentum is ${\vec p _f}= m{\vec v_f}$. So Eq. \eqref{2} can be rewritten as,

\[\vec J = {\vec p _f} - {\vec p _i} \tag{3} \label{3}\]

We can also express Newton's Second law as the rate of change of linear momentum. Newton's second law gives

\[\begin{align*} {\vec F} &= m\vec a \\ &= m\frac{{{{\vec v }_f} - {{\vec v }_i}}}{{\Delta t}}\\ &= \frac{{m{{\vec v }_f} - m{{\vec v }_i}}}{{\Delta t}}\\ {\rm{or,}}\quad {\vec F} &= \frac{{{{\vec p }_f} - {{\vec p }_i}}}{{\Delta t}} \tag{4} \label{4} \end{align*}\]

In terms of varying force we can write the force in a very small interval of time $dt$ in which $\Delta t$ in Eq. \eqref{4} approaches zero and $d\vec p = {\vec p _f} - {\vec p _i}$.

\[{\vec F} = \frac{{d\vec p }}{{dt}} \tag{5} \label{5}\]

We can show that Eq. \eqref{3} is true even if the force is not constant. We first find the impulse in infinitesimally small time interval $dt$ and integrate it from the initial linear momentum ${p _i}$ to the final linear momentum ${p _f}$ as

\[J = \int\limits_{{{p }_i}}^{{{ p }_f}} {\frac{{dp }}{{dt}}dt} = {p _f} - {p _i}\]

Which is the same as Eq. \eqref{3} without vector sign. Of course, you can put the vector arrow to indicate it as a vector quantity.