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#### Definition of Kinematics And Its Equations

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### Definition of Kinematics

Kinematics is the branch of classical mechanics, which deals with moving objects. In kinematics, the forces that cause the motion are not important. It is also referred to as the feature of the motion in any object. In the field of study, Kinematics is often mentioned to as “geometry of motion”, and can be also seen as a branch of mathematics. A problem of kinematics often starts with explaining the geometry of the system and it continues with declaring the primary conditions for any values of a position that are known, and the velocity or the acceleration of points within the system. After that, using those arguments from the geometry, position, the velocity and the acceleration, for any unknown part of the system it is determined. The study of forces and the how they act on a body is known as kinetics.

### History of the word

The word “kinematic” is derived from the Greek word kinema, which means “movement” or “motion”, which in turn is derived from the word, kinein, the meaning of which is “to move”.

The fields where kinematics is used

Kinematics is used in many fields. In astrophysics, kinematics helps to define the motion of the celestial bodies. In the field of mechanical engineering, robotics and biomechanics, the motion of the systems, which are multi-link systems, are described using kinematics. Examples of multi-link systems can be, engines, robotics arms and the human skeleton. In mechanical system, geometric transformations, also known as rigid transformations, are used to simplify the derivation for the equations of the motion. The geometric transformation is also important to dynamic analysis.

The process to measure kinematic quantities that are used to describe the motion, is known as kinematic analysis. In the field of engineering, kinematic analysis is used to in order to find the range for a movement in a given mechanism. In order to design the mechanism for any given range of motion, kinematic synthesis is used.

### Formulas that are used in kinematic

In a kinematic problem, there are mainly five variable that are used which are:

1. Δx denotes Displacement

2. t denotes Time interval

3. v0 ​is used to mention the Initial velocity

4. v is used to denote Final velocity and finally,

5. a is used to mention the Constant acceleration.

There are mainly four formulae that are used in kinematic problem which are:

How to derive the first formula of kinematic:

The first formula of kinematic is the easiest of all to derive because with some rearrangement of the definition of the acceleration it can be done. In the starting, the definition of acceleration is, a = Δv​ / Δt

Now, the Δv is replaced in the definition with change in velocity v – v0. So, the equation becomes,

a = v – v0 / Δt

Finally, just to solve v,

v= v0​ + aΔt

If in the formula in place of Δt only t is used, the first formula can be derived,

v = v0​ + at

The strategy to solve the kinetic problems:

Identification to use the correct equation in the right place is the main strategy to solve a problem. The following steps should be followed:

1. Firstly, the construction of an informative diagram is important, for the physical situation.

2. Identification and listing of the given information of the variables.

3. Identification and listing of the unknown information.

4. Identification and listing of the equations that should be used in solve the problem.

5. Substitution of the known values to the equation and after that use of correct algebraic steps in order to find the unknown information.

6. The answer should be checked for any mathematical errors and if the answer is correct.

Example problems

1. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

Solution:

Given: a = + 3.2 m/s2 , t = 32.8 s and vi = 0 m/s

d = vi*t + 0.5*a*t2

d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2

d = 1720 m

2. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.

Solution:

Given: d = 110 m, t = 5.21 s and vi = 0 m/s

d = vi*t + 0.5*a*t2

110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2

110 m = (13.57 s2)*a

a = (110 m)/(13.57 s2)

a = 8.10 m/ s2

3. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.

Solution:

Given:

 vi = 18.5 m/s vf = 46.1 m/s t = 2.47 s

Find:

d = ??

a = ??

a = (Delta v)/t

a = (46.1 m/s - 18.5 m/s) / (2.47 s)

a = 11.2 m/s2

d = vi*t + 0.5*a*t2

d = (18.5 m/s) * (2.47 s)+ 0.5*(11.2 m/s2) * (2.47 s)2

d = 45.7 m + 34.1 m

d = 79.8 m

(Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)

4. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.83 seconds, then determine the acceleration and to what distance the sled travels?

Solution:

Given: vi = 0 m/s, vf = 444 m/s and t = 1.83 s

Find: a = ??

d = ??

a = (Delta v)/t

a = (444 m/s - 0 m/s)/(1.83 s)

a = 243 m/s2

d = vi*t + 0.5*a*t2

d = (0 m/s)*(1.83 s)+ 0.5*(243 m/s2)*(1.83 s)2

d = 0 m + 406 m

d = 406 m

(Note: the d can also be calculated using the equation vf2 = vi2 + 2*a*d)

Find: a = ??

Find: d = ??

OR

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