Position → Velocity → Acceleration → Force

The velocity of a particle whose motion is described by position vector $\vec{r}$ is the derivative of the position vector with respect to time $t$,

$$\vec{v}=\frac{d\vec{r}}{dt}.$$

To differentiate a vector, take the derivative of all three components of the vector,

$$\vec{v}= \frac{dr_x}{dt}\hat{x}+ \frac{dr_y}{dt}\hat{y}+ \frac{dr_z}{dt}\hat{z}.$$

The acceleration is the derivative of the velocity with respect to time,

$$\vec{a}=\frac{d\vec{v}}{dt}.$$

The force on the particle is related to the acceleration by Newton's law,

$$\vec{F} = m\vec{a},$$

here $m$ is the mass of the particle.

In summary, starting with the time dependent position vector that describes the path that a particle follows in three dimensions, the velocity, acceleration, and force on this particle can be calculated.

$$ \vec{r}(t)\text{ [m]},$$ $$ \vec{v}(t)=\frac{ d\vec{r}}{dt}\text{ [m/s]},$$

$$\vec{a}=\frac{ d^2\vec{r}}{dt^2} = \frac{ d\vec{v}}{dt}\,[\text{ m/s}^2],$$ $$\vec{F}=m\frac{ d^2\vec{r}}{dt^2} = m\frac{ d\vec{v}}{dt}=m\vec{a}\text{ [N]}. $$

$\vec{r}(t)=$  $\hat{x}$ +  $\hat{y}$ +  $\hat{z}$  

$\vec{v}(t) = $ () $\hat{x}$ + () $\hat{y}$ + () $\hat{z}$

$\vec{a}(t) = $ () $\hat{x}$ + () $\hat{y}$ + () $\hat{z}$

Questions