### Theory:

Consider an object of mass, $$m$$ with an initial velocity, $$u$$ is moving along a straight line. It is in uniform acceleration with velocity, $$v$$ in time, $$t$$ by a constant force, $$F$$.

Initial momentum of the object, $$p_1$$ $$=$$ $$m\ u$$
Final momentum of the object, $$p_2$$ $$=$$ $$m\ v$$

$\begin{array}{l}\mathit{Change}\phantom{\rule{0.147em}{0ex}}\mathit{in}\phantom{\rule{0.147em}{0ex}}\mathit{momentum}\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}{p}_{2}-{p}_{1}\\ \mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\mathit{mv}-\mathit{mu}\\ \mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}m×\left(v-u\right)\\ \mathit{Rate}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{change}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{momentum}\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\frac{m×\left(v-u\right)}{t}\end{array}$

$\begin{array}{l}F\phantom{\rule{0.147em}{0ex}}\mathrm{\alpha }\phantom{\rule{0.147em}{0ex}}\phantom{\rule{0.147em}{0ex}}\frac{m×\left(v-u\right)}{t}\\ \\ F\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{\mathit{km}×\left(v-u\right)}{t}\end{array}$

Where $$k$$ is the constant of proportionality.

$F\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}k\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}a$

Here, $a\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{\left(v-u\right)}{t}$

The above equation, which implies the rate of change of velocity, is known as acceleration.
Unit of force:
We know that the SI units of mass is $$kg$$ and acceleration is $$m\ s^{ –2}$$. The unit of force is selected in such a way that the value of the proportionality constant, $$k$$ is one.
A unit of force is known as the amount of force that causes an acceleration of $$1\ m\ s^{ –2}$$ in an object having $$1\ kg$$ mass.

Hence, the value of $$k$$ becomes one and makes $F\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}a$.

The SI unit of force is $$kg\ m\ s^{ –2}$$ or newton ($$N$$). The value of one newton is written as,
$$1\ N$$ $$=$$ $$1\ kg\ m\ s^{ –2}$$
The second law of motion implies that the force ($$F$$) acting on an object is a product of its mass ($$m$$) and acceleration ($$a$$).

$F\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}m\phantom{\rule{0.147em}{0ex}}a$