This article focuses on kinematics formulas and their derivation. Kinematics refers to the study of the motion of points, objects, and group of objects while ignoring the causes of its motion. Kinematics refers to the branch of classical mechanics which describes the motion of points, objects, and systems comprising of groups of objects. Some experts refer to the study of kinematics as the “geometry of motion”.
What is Kinematics?
Simply speaking, kinematics refers to the study of motion. Kinematics certainly deals with any type of motion of any particular object. Kinematics refers to the study of objects in motion as well as their inter-relationships. Furthermore, kinematics is a branch of classical mechanics and it explains and describes the motion of points, objects, and systems of bodies.
In order to describe the motion, kinematics focuses on the trajectories of points, lines, and various other geometric objects. Moreover, it also focuses on the various deferential properties including velocity and acceleration. Also, there is heavy usage of kinematics in astrophysics, mechanical engineering, robotics, and biomechanics.
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Kinematics Formulas
There are four kinematics formulas and they relate to displacement, velocity, time, and acceleration. Furthermore, the four kinematic formulas are as follows:
1. A = \(\frac{v_{f} – v_{i}}{t}\)
2. \(\frac{v_{i} + v_{f}}{2}\) = \(\frac{D}{t}\)
3. D = vit + \(\frac{1}{2}at^{2}\)
4. vf^{2} = vi^{2} + 2Ad
Where,
D = displacement
a = acceleration
t = time
vf = final velocity
vi = initial velocity
Kinematics Formulas Derivations
First of all, one must calculate the slope of the diagonal line. Here, the slope would be a change in velocity and divided by a change in time. Furthermore, the slope would equal the acceleration.
a = \(\frac{v_{2} – v_{1}}{t_{2} – t_{1}}\)
One must rewrite t2 – t1 as \(\Delta t\)
a = \(\frac{v_{2}-v_{1}}{\Delta t}\). This is certainly equation 1. One must rearrange it so as to get v2 on the left side. This would certainly express the formula in the slope-intercept form of a line.
v2 = v1 + a\(\Delta t\)
To get the next formula, one must first derive an expression for the displacement of the object. Furthermore, the time interval is \(\Delta t\). The calculation of the displacement is below:
S = v\(\Delta t\)
Furthermore, the displacement of the object is certainly equal to v1\(\Delta t\) . Also, the product v1 is equal to the area A1.
So A1 = v1\(\Delta t\)
Then, A2 = \(\frac{1}{2}(V_{2}-V_{1}\Delta t)\)
Now adding A1 and A2
s = A1 + A2
Substituting for A1 and A2 gives
s = \(\frac{1}{2}\left ( v_{2}-v_{1} \right )\Delta t\) + \(v_{1}\Delta t\)
Now simplifying it would give
s = \(\frac{1}{2}\left ( v_{2}+v_{1} \right )\Delta t\). This is the equation 2.
Equation no 3 is found out by eliminating v2
One must start with formula 1
v2 = v1 + a\(\Delta t\)
Now one must apply some algebra so as to make the left side of the formula to look like the right side of formula 2
v2 + v1 = v1 + a\(\Delta t\)+ v1
v2+ v1 = 2v1 + a\(\Delta t\)
Furthermore, one must multiply both the sides by \(\frac{1}{2}\Delta t\)
s = \(\frac{1}{2}\left ( v_{2}+v_{1} \right )\Delta t\) = \(\frac{1}{2}\left ( 2v_{1} + a\Delta t \right )\Delta t\)
s = \(v_{1}\Delta t + \frac{1}{2}a\Delta t^{2}\). This is formula 3
Formula 4 is found by eliminating the time variable, or \(\Delta t\)
Now, one must certainly begin with equation 1 whose rearrangement has taken place with the acceleration on the left side of the equals sign
a = \(\frac{v_{2}-v_{1}}{\Delta t}\)
Furthermore, one must multiply the left side of equation 1 by the left side of equation 2. Moreover, one must multiply the right side of equation 1 with the right side of equation 2.
s = \(\frac{1}{2}\left ( v_{2}+v_{1} \right )\Delta t\)
as = \(\left [ \frac{1}{2}\left ( v_{2} – v_{1}\right )\Delta t \right ]\left [ \frac{v_{2}-v_{1}}{\Delta t} \right ]\)
Then \(\Delta t\) cancels out which certainly leads to the simplification of the equation.
2as = \(v2^{2}-v1^{1}\)
This formula is almost always written as:
\(v2^{2}\)= \(v1^{1}\) + 2as. This is formula 4.
Solved Examples
Q1. A man is riding his bicycle to the store at a velocity of 4 m/s. Then, suddenly a cat runs out in front of him. He rapidly brakes to a complete stop, with an acceleration of – 2m/s2. Find out his displacement?
Answer: Here, the final velocity, vf = o. Also, the initial velocity, vi = 4m/s. Furthermore, the acceleration, a = -2m/s2.. Therefore, one must use the following formula:
\(vf^{2}\) = \(vi^{2}\) + 2Ad
0 = (4)2 +2(-2)D
0 = 16 + (-4)D
16 = (4)D
D = 16/4 = 4
Hence, the total displacement is 4m.
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