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The goal of this first unit that is of the Physics classroom has been to investigate the variety which is of means by which the motion of objects can be described. The variety of representations that we have already learnt and investigated also includes verbal representations that are pictorial representations as well as the numerical representations and graphical representations that are the position-time graphs and velocity-time graphs.

The branch which is of physics that generally defines motion which is with respect to time and space that is ignoring the cause of that motion is known as kinematics. The equation which is the Kinematics is a set of equations that can derive an unknown aspect of motions of the bodyâ€™s if the other aspects are provided.

These equations link five kinematic variables:

The Displacement that is denoted by Î”x.

The initial velocity that is given as v

_{0}which is said to be a Final velocity that is denoted by a v Time interval that is denoted by t Constant acceleration that is denoted by a.

Essentially we can say that the kinematics equations which can derive one or more of these variables if the others are given. These equations which we have seen define motion at either constant velocity or at constant acceleration. Because we can say that the kinematics equations are only applicable at a constant acceleration or a speed which is constant we cannot use them if there is either of the two which is changing.

The knowledge of each of the different quantities generally provides descriptive information about an object's motion. For example, we can say that if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement that is of 264 meters. Then we can say that the motion of the car is fully described.

If we take a second car then this is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s^{2} for a time of 8.0 seconds, and the providing which is a final velocity of 24 m/s. The east and an eastward direction of displacement is 96 meters and then we can say that the motion of this car is fully described.

These two statements which we have seen is to provide a complete description of the motion of an object. However, we can say that such completeness is not always known.

It is often seen that only a few parameters of an object's motion are known while the rest are unknown. For example, we can say that as we approach the stoplight then we might know that the car has a velocity of 22 m/s, is moving in the east direction and is capable of a skidding acceleration of 8.0 m/s^{2} and to the direction which is west. However, we do not know the displacement that our car would experience if we were to slam on your brakes and skid to a stop. and then after that, we do not know the time required to skid to a stop. In such an instance like this, we can say that the unknown parameters that can be determined using physics principles and mathematical equations are the kinematic equations.

The equations that can be utilized for any motion that can be generally described as being either a constant motion which is of velocity motion that is an acceleration of 0 m/s/s or we can say a constant acceleration motion. They can be said to be never used over any time period during which the acceleration is changing. Each of the equations which are the kinematic equations includes four variables. If the values of three of the four variables are known to us then the value of the fourth variable can just be calculated.

In this manner, we can say that the equation which is the kinematic equations provide a useful means of predicting information about an object's motion if other information is already known to us. For example, we can say that if the acceleration value and the initial and final which is of the velocity values of a skidding car is known then we can say that the displacement of the car and the time can be predicted using the kinematic equations. Here, we will focus upon the use of the kinematic equations that is to predict the numerical values of unknown quantities for objects in motion.

The four kinematic equations that describe an object's motion are given below:

\[d = v_{i} \times t + \frac{1}{2} \times a \times t^{2}\] \[v_{f}^{2} = v_{i}^{2} + 2 \times a \times d\]

\[v_{f} = v_{i} + a \times t\] \[d = \frac{v_{i} + v_{f}}{2} \times t\]

There are a variety of symbols that are used in the above equations that we have seen. Each symbol that we have seen has its own specific meaning. d stands for the displacement of the object, t stands for the time for which the object moved, a stands for the acceleration of the object, v stands for the velocity of the object, v, indicates that the velocity value is the initial while vf indicates that the value of the velocity that is final.

Each of these four equations that we have seen appropriately describes the mathematical relationship which is between the parameters of an object's motion. As such we can say they can be used to predict unknown information which is about an object's motion that is if other information is known. In the next part we will investigate the process of doing this.

FAQ (Frequently Asked Questions)

Q1. What are the 5 Parameters in Kinematic Equations?

Ans: Building which is on what we have learned so far and what Galileo presented, in kinematic equations the letter v is said to be a velocity, the letter x is position and t is time and a is acceleration. Remember the symbol Î” means a change in entropy.

Q2. How can Kinematic Equations be Paired up?

Ans: Our goal which is in this section then we can say is to derive new equations that can be used to describe the motion that is of an object in terms of its three kinematic variables: the velocity denoted by v the position letter s devoted it and time t. There are three ways which to pair them up: that is the velocity-time, and then the position-time, and the last velocity-position.

Q3. How do you Calculate Kinematics?

Ans: The kinematic formula which is denoted as Î”x = v_0t + Â½ at^2Â Î”x = v_0 t+ Â½ at^2 Î”x = v_0t + 1 at 2 delta, x, equals, v that start subscript which is of 0, that end subscript which is of letter t, plus, start fraction, 1, divided by, 2, end fraction, a, t, squared which is missing letter v, so it's the right choice which is, in this case, to solve for the acceleration denoted by letter a.