Equations for Kinematics – Formulae, Derivation
What is Kinematics?
Kinematics is the study of moving objects. Kinematics is concerned with any form of motion of any specific object. Kinematics is the study of moving objects and their interrelationships. In addition, kinematics is a part of classical mechanics that describes and explains the motion of points, objects, and systems of things.
Kinematics focuses on the trajectories of points, lines, and other geometric objects to explain motion. Furthermore, it concentrates on numerous deferential qualities like velocity and acceleration. Kinematics is also widely used in astronomy, mechanical engineering, robotics, and biomechanics.
What are the kinematic formulas?
The kinematic formulae are a collection of formulas that connect the five kinematic variables mentioned below.
\(\Delta x\) is Displacement
\(t\) is Time interval
\(v_{0} \) Initial velocity
\(v\) Final velocity
\(a\) Constant acceleration
If we know three of the five kinematic variables \(\Delta x\), \(t\), \(v_{0} \), \(v\), \(a\)
for an object under constant acceleration, we can solve for one of the unknown variables using the kinematic formula.
The kinematic formulae are frequently stated as the four equations below.
- \(v= v_{0}+at\)
- \(\Delta x=(v+ v_{0})t\)
What is Inverse Kinematics?
Inverse Kinematics is the reverse of kinematics, and if we know the endpoint of a certain structure, we may calculate the angle values required for the joints to reach that endpoint. It’s a little challenging because there are usually several, if not infinite ways.
\(v= v_{0} +at\)
\(\Delta x=(v+ \frac{ v_{0} }{2} )t\)
\(\Delta x= v_{0} t+ \frac{1}{2} a t^{2}\)
\(v^{2} = v_{0} ^{2} +2a \Delta x\)
We can determine the fifth variable using kinematic equations if any four of the variables are known.
What is Rotational Kinematics?
We have been studying the Translational or linear kinematics equation, which deals with the motion of a linearly moving body. Another area of kinematics equations deals with any person’s rotating motion. These, however, are just corollaries of the previous equations with only the variables modified.
- A change in angle replaces displacement.
- Initial and final angular velocity replace initial and final velocity.
- Angular acceleration replaces acceleration.
- The only constant is time.
Rotational Motion, \(\alpha \) is constant
\(w= w_{0} + \alpha t\)
\(\Theta = \frac{1}{2} (w+ w_{0} )t\)
\(\Theta = w_{0}t+ \frac{1}{2} \alpha t^{2}\)
\(w^{2} = w^{2} _{0} +2 \alpha \Theta\)
Linear Motion (a = constant)
\(v= v_{0} +at\)
\(x=(v+ \frac{ v_{0} }{2} )t\)
\(x= v_{0} t+ \frac{1}{2} a t^{2}\)
\(v^{2} = v_{0} ^{2} +2a x\)
Derivations of Kinematics
First, the slope of the diagonal line must be calculated. The slope in this case would be a change in velocity divided by a change in time. In addition, the slope would match the acceleration.
\(a= \frac{ v_{2} – v_{1} }{ t_{2} – t_{1} }\)
Rewriting, \(t_{2} – t_{1}= \Delta t\)
Then,
\(a= \frac{ v_{2} – v_{1} }\Delta t\)
This is undoubtedly equation 1. It must be rearranged such that v2 is on the left side. This would express the formula as a slope intercept in a line.
\(v_{2} = v_{1} +a \Delta t\)
To acquire the next formula, one must first derive an equation for the object’s displacement. In addition, the time interval is t. The displacement is calculated as follows:
\(S=v \Delta t\)
Furthermore, the object’s displacement is unquestionably equal to v1t. Furthermore, product v1 equals area A1.
\(A_{1} = v_{1} \Delta t\)
Then,
\(A_{2} = \frac{1}{2} ( v_{2}- v_{1} \Delta t )\)
Now adding A1 and A2
\(s= A_{1} + A_{2}\)
Substituting for A1 and A2 gives
\(s= \frac{1}{2} ( v_{2} – v_{1} ) \Delta t+ v_{1} \Delta t\)
Now simplifying it would give,
\(s= \frac{1}{2} ( v_{2} + v_{1} ) \Delta t\)
And this is Formula 2
We get equation no 3 by eliminating v2
One must start with formula 1,
\(v_{2} = v_{1}+a \Delta t\)
Now, some algebra is required to make the left half of the formula seem like the right side of formula 2.
\(v_{2} + v_{1} = v_{1}+a \Delta t+ v_{1}\)
\(v_{2} + v_{1}=2 v_{1} +a \Delta t\)
Furthermore, one must multiply both sides by, \(\frac{1}{2} \Delta t\)
\(s= \frac{1}{2} ( v_{2} + v_{1} ) \Delta t= \frac{1}{2} (2 v_{1}+a \Delta t ) \Delta t \)
\(s= v_{1} \Delta t+ \frac{1}{2} a \Delta t^{2}\)
This is Formula 3.
Formula 4 is obtained by removing the time variable, or Δt.
Now, one must undoubtedly start with equation 1, which has been rearranged with the acceleration on the left side of the equal’s sign.
\(a= \frac{v_{2} – v_{1} }{ \Delta t}\)
Furthermore, the left side of equation 1 must be multiplied by the left side of equation 2. Furthermore, the right side of equation 1 must be multiplied by the right side of equation 2.
\(s= \frac{1}{2} ( v_{2}+ v_{1} ) \Delta t\)
\(as=[ \frac{1}{2}( v_{2} – v_{1} ) \Delta t ][ \frac{v_{2} – v_{1}}{ \Delta t} ]\)
Then Δt cancels out which certainly leads to the simplification of the equation.
\(2as= v_{2} ^{2} -v_{1} ^{2}\)
This formula is almost always written as:
\(v_{2} ^{2}=v_{1} ^{2}+2as\)
Final Notes
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