# Question #c26f8

Jun 5, 2016

I've done it below, check it.

#### Explanation:

From Newton's second law, we know $\vec{F} = \frac{\mathrm{dv} e c p}{\mathrm{dt}}$

Where $\vec{p}$ is the linear momentum.

Now, by definition of momentum, $\vec{p} = m \vec{v}$ where, $m$ is the mass and $\vec{v}$ is the velocity.

We thus have, $\vec{F} = \frac{d}{\mathrm{dt}} \left(m \vec{v}\right)$

For a constant mass $m$, $\vec{F} = m \frac{\mathrm{dv} e c v}{\mathrm{dt}}$

By definition of acceleration, $\vec{a} = \frac{\mathrm{dv} e c v}{\mathrm{dt}}$, where $\vec{a}$ is the acceleration.

We therefore have the result, $\vec{F} = m \vec{a}$.

Now, for a zero external force, $\vec{F} = 0$ which implies, $m \vec{a} = 0$

For a non zero mass, we have $\vec{a} = 0$. This implies that on zero external net force being applied, the acceleration is $0$.

In other words, when no (net) external force acts on the body, if it is in rest, it continues to be in it's state of rest or if it is moving along a straight line, it continues to move along it neither gaining nor losing speed.

This is the Newton's first law.