Editing Kinematics (section)

===Relative acceleration===

The acceleration of one point ''C'' relative to another point ''B'' is simply the difference between their accelerations.
<math display="block">\mathbf{a}_{C/B} = \mathbf{a}_{C} - \mathbf{a}_{B} </math>
which is the difference between the components of their accelerations.

If point ''C'' has acceleration components <math>\mathbf{a}_{C} = \left( a_{C_x}, a_{C_y}, a_{C_z} \right) </math>
and point ''B'' has acceleration components <math>\mathbf{a}_{B} = \left( a_{B_x}, a_{B_y}, a_{B_z} \right) </math>
then the acceleration of point ''C'' relative to point ''B'' is the difference between their components: <math>\mathbf{a}_{C/B} = \mathbf{a}_{C} - \mathbf{a}_{B} = \left( a_{C_x} - a_{B_x} , a_{C_y} - a_{B_y} , a_{C_z} - a_{B_z} \right) </math>

Assuming that the initial conditions of the position, <math>\mathbf{r}_0</math>, and velocity <math>\mathbf{v}_0</math> at time <math>t = 0</math> are known, the first integration yields the velocity of the particle as a function of time.<ref>{{Citation | title=2.4 Integration | date=2 June 2017 | url=https://www.youtube.com/watch?v=H7xmTMQ265s | archive-url=https://ghostarchive.org/varchive/youtube/20211113/H7xmTMQ265s| archive-date=2021-11-13 | url-status=live| publisher=MIT | language=en | access-date=2021-07-04}}{{cbignore}}</ref>
<math display="block">\mathbf{v}(t) = \mathbf{v}_0 + \int_0^t \mathbf{a}(\tau) \, \text{d}\tau</math>

Additional relations between displacement, velocity, acceleration, and time can be derived. If the acceleration is constant,
<math display="block">\mathbf{a} = \frac{\Delta\mathbf{v}}{\Delta t} = \frac{\mathbf{v}-\mathbf{v}_0}{ t } </math> can be substituted into the above equation to give:
<math display="block">\mathbf{r}(t) = \mathbf{r}_0 + \left(\frac{\mathbf{v} + \mathbf{v}_0}{2}\right) t .</math>

A relationship between velocity, position and acceleration without explicit time dependence can be obtained by solving the average acceleration for time and substituting and simplifying

<math display="block"> t = \frac{\mathbf{v}-\mathbf{v}_0}{ \mathbf{a} } </math>

<math display="block"> \left(\mathbf{r} - \mathbf{r}_0\right) \cdot \mathbf{a} = \left( \mathbf{v} - \mathbf{v}_0 \right) \cdot \frac{\mathbf{v} + \mathbf{v}_0}{2} \ , </math>
where <math> \cdot </math> denotes the [[dot product]], which is appropriate as the products are scalars rather than vectors.
<math display="block">2 \left(\mathbf{r} - \mathbf{r}_0\right) \cdot \mathbf{a} = |\mathbf{v}|^2 - |\mathbf{v}_0|^2.</math>

The dot product can be replaced by the cosine of the angle {{mvar|α}} between the vectors (see [[Dot product#Geometric definition|Geometric interpretation of the dot product]] for more details) and the vectors by their magnitudes, in which case:
<math display="block">2 \left|\mathbf{r} - \mathbf{r}_0\right| \left|\mathbf{a}\right| \cos \alpha = |\mathbf{v}|^2 - |\mathbf{v}_0|^2.</math>

In the case of acceleration always in the direction of the motion and the direction of motion should be in positive or negative, the angle between the vectors ({{mvar|α}}) is 0, so <math>\cos 0 = 1</math>, and
<math display="block"> |\mathbf{v}|^2= |\mathbf{v}_0|^2 + 2 \left|\mathbf{a}\right| \left|\mathbf{r}-\mathbf{r}_0\right|.</math>
This can be simplified using the notation for the magnitudes of the vectors <math>|\mathbf{a}|=a, |\mathbf{v}|=v, |\mathbf{r}-\mathbf{r}_0| = \Delta r </math>{{citation needed|date=April 2018}} where <math>\Delta r</math> can be any curvaceous path taken as the constant tangential acceleration is applied along that path{{citation needed|date=April 2018}}, so
<math display="block"> v^2= v_0^2 + 2a \Delta r.</math>
This reduces the parametric equations of motion of the particle to a Cartesian relationship of speed versus position. This relation is useful when time is unknown. We also know that <math display="inline">\Delta r = \int v \, \text{d}t</math> or <math>\Delta r</math> is the area under a velocity–time graph.<ref>https://www.youtube.com/watch?v=jLJLXka2wEM Crash course physics integrals</ref> [[File:Velocity Time physics graph.svg|thumb|Velocity Time physics graph]] We can take <math>\Delta r</math> by adding the top area and the bottom area. The bottom area is a rectangle, and the area of a rectangle is the <math>A \cdot B</math> where <math>A</math> is the width and <math>B</math> is the height. In this case <math>A = t</math> and <math>B = v_0</math> (the <math>A</math> here is different from the acceleration <math>a</math>). This means that the bottom area is <math>tv_0</math>. Now let's find the top area (a triangle). The area of a triangle is <math display="inline">\frac{1}{2} BH</math> where <math>B</math> is the base and <math>H</math> is the height.<ref>https://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html Area of Triangles Without Right Angles</ref> In this case, <math>B = t</math> and <math>H = at</math> or <math display="inline">A = \frac{1}{2} BH = \frac{1}{2} att = \frac{1}{2} at^2 = \frac{at^2}{2}</math>. Adding <math>v_0 t</math> and <math display="inline">\frac{at^2}{2}</math> results in the equation <math>\Delta r</math> results in the equation <math display="inline">\Delta r = v_0 t + \frac{at^2}{2}</math>.<ref>{{Cite AV media |url=https://www4.uwsp.edu/physastr/kmenning/Phys203/eqs/kinematics.gif |title=kinematics.gif (508×368) |type=Image |language=en |access-date=3 November 2023}}</ref> This equation is applicable when the final velocity {{mvar|v}} is unknown.

[[File:Nonuniform circular motion.svg|thumb|250px|Figure 2: Velocity and acceleration for nonuniform circular motion: the velocity vector is tangential to the orbit, but the acceleration vector is not radially inward because of its tangential component '''a'''<sub>''θ''</sub> that increases the rate of rotation: d''ω''/d''t'' = <nowiki>|</nowiki>'''a'''<sub>''θ''</sub><nowiki>|</nowiki>/''R''.]]