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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

r=xx^+yy^+zz^v=r˙=x˙x^+y˙y^+z˙z^a=v˙=r¨=x¨x^+y¨y^+z¨z^ \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 r\mathbf{r} with respect to tt yields the following.

v=ddt(xx^+yy^+zz^)=x˙x^+xx^˙+y˙y^+yy^˙+z˙z^+zz^˙ \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 x^˙=y^˙=z^˙=0\dot{\hat{\mathbf{x}}}=\dot{\hat{\mathbf{y}}}=\dot{\hat{\mathbf{z}}} = 0, therefore, it follows that:

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

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

Acceleration

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

a=ddt(x˙x^+y˙y^+z˙z^)=x¨x^+y¨y^+z¨z^ \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