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Velocity and Acceleration in Cartesian Coordinate System 📂Classical Mechanics

Velocity and Acceleration in Cartesian Coordinate System

Velocity and Acceleration in Cartesian Coordinates

$$ \begin{align*} \mathbf{r} &= x \hat{\mathbf{x}} + y \hat{\mathbf{y}} + z \hat{\mathbf{z}} \\ \mathbf{v} &= \dot{\mathbf{r}} = \dot{x} \hat{\mathbf{x}} + \dot{y} \hat{\mathbf{y}} + \dot{z} \hat{\mathbf{z}} \\ \mathbf{a} &= \dot{\mathbf{v}} = \ddot{\mathbf{r}} = \ddot{x} \hat{\mathbf{x}} + \ddot{y} \hat{\mathbf{y}} +\ddot{z}\hat{\mathbf{z}} \end{align*} $$

Derivation

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Determining velocity and acceleration in a Cartesian coordinate system is straightforward.

Velocity

Differentiating $\mathbf{r}$ with respect to $t$ yields the following.

$$ \mathbf{v}=\frac{d}{dt}(x\hat{\mathbf{x}} +y\hat{\mathbf{y}}+z\hat{\mathbf{z}})=\dot{x} \hat{\mathbf{x}} + x\dot{\hat{\mathbf{x}}}+\dot{y} \hat{\mathbf{y}} + y\dot{\hat{\mathbf{y}}} +\dot{z} \hat{\mathbf{z}} + z\dot{\hat{\mathbf{z}}} $$

Since the unit vectors of the Cartesian coordinate system are independent of time changes, they are $\dot{\hat{\mathbf{x}}}=\dot{\hat{\mathbf{y}}}=\dot{\hat{\mathbf{z}}} = 0$, therefore, it follows that:

$$ \mathbf{v} = \dot{x} \hat{\mathbf{x}} + \dot{y} \hat{\mathbf{y}} + \dot{z} \hat{\mathbf{z}} $$

Notably, $\dot{r}$ is read as [al dot]. In physics, a dot over a letter signifies differentiation with respect to time.

Acceleration

Differentiating $\mathbf{v}$ with respect to $t$ results in the following.

$$ \mathbf{a}=\frac{d}{dt}(\dot{x} \hat{\mathbf{x}}+\dot{y} \hat{\mathbf{y}}+\dot{z} \hat{\mathbf{z}})=\ddot{x} \hat{\mathbf{x}}+\ddot{y} \hat{\mathbf{y}}+\ddot{z}\hat{\mathbf{z}} $$

See also