Position, Velocity and Acceleration Vectors [Course Physics1]

The position of a mobile at a time t is determined with respect to a reference frame by a vector $\vec{OM}$ which is called the position vector. Its origin is the center of the frame O and its end is the mobile M^[1]

$\vec{OM} = \vec{OM} (t)$

We define the vector $\vec{M_{1} M_{2}}$ the displacement vector

$\vec{M_{1} M_{2}} = \vec{{OM}_{2}} - \vec{{OM}_{1}}$

The average speed is defined as follows

$\vec{V_{a}} = \frac{\vec{{OM}_{2}} - \vec{{OM}_{1}}}{t_{2} - t_{1}} = \frac{\vec{M_{1} M_{2}}}{Δ t}$

It is the velocity at a given time t and it is defined as follows

$\vec{V} = \lim_{Δ t \to 0} \vec{V_{a}} = \frac{d \vec{OM}}{dt}$

The instantaneous velocity vector is tangent to the trajectory and its direction follows the direction of movement.

Average acceleration is the change in speed between two positions with respect to time. The mobile undergoes an average acceleration such that

$\vec{a_{a}} = \frac{\vec{V_{2}} - \vec{V_{1}}}{t_{2} - t_{1}} = \frac{Δ \vec{V}}{Δ t}$

$V_{1}$ is the speed of the mobile at time $t_{1}$ and $V_{2}$ its speed at time $t_{2}$

Instantaneous acceleration is the acceleration at a given time t

$\vec{a} = \lim_{Δ t \to 0} \vec{a_{a}} = \frac{d \vec{V}}{dt} = \frac{d^{2} \vec{OM}}{{dt}^{2}}$