Equations of kinematics in two dimensions - العلم نور

# العلم نور

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## Equations of kinematics in two dimensions Equations of kinematics in two dimensions

### Kinematics

It is one of classical mechanics branches which concerned with the description of objects’ motion without considering the forces acting on them to move. Kinematics equations are used for calculating the missing variables of any system such as position, acceleration or velocity of the moving body.
The most motions in our world are in three dimensions but two dimensions are used to simplify the analysis of many cases of motion. In this article, we are willing to discuss the equations of kinematics in two dimensions.

### Kinematics in Two Dimensions

Tow dimensional motion is the motion of a body which is thrown in the air at an angle. This body motion will appear as a curve in the space as described below: Kinematics in Two Dimensions

In two-dimensional motion, the object seems to move in both vertical and horizontal directions throw the space. In this motion, only the vertical velocity will change during the motion and the horizontal velocity remains constant along the motion path.

### Equations of Kinematics in Two Dimensions

In order to understand the equations of kinematics in two dimensions, it’s required to analyze the motion separately in the horizontal direction (x) and the vertical direction (y). There are different sets of equations to describe each direction of motion.

For Horizontal Direction:
As we mentioned above, the velocity in this direction remains constant along the path of motion so, there is no acceleration in this direction. Also, the air resistance is considered to be neglected so, the following equation is used for horizontal motion:
Where,
-        ΔX: is the distance which the body moved.
-        vx: is the constant velocity in the horizontal direction.
-        t: is the time of motion.

For Vertical Direction:
The vertical motion of the body will be under the influence of gravity so, the acceleration in this case will be constant and equals 9.8 m/sec2. The equations in the vertical dimension will be as follows:
vy ​= v0y ​+ ayt
vy2 ​= v0y2​ + 2ayΔy
Where,
-        v0y: is the initial velocity of the motion in y direction (m/s).
-        vy: is the final velocity of the motion in y direction (m/s).
-        Δy: is the altitude from which the object is falling (m).
-        t: is the time of the motion (sec.).
-        ay: is the acceleration of the motion in vertical direction, ay = 9.8 m/sec2.

### Problem Solving

In order to solve any problem of the kinematics in two dimensions, you should know at first what equation you will use according two the given data in the problem. After that, you will be able to calculate the required variables easily.

Example: A ball is thrown horizontally with a vx of 4 m/s from the third floor of a building of altitude equals 12 m. How far does the ball travel horizontally before striking the ground?
Solution
-        At first, we will use the equations of vertical direction to calculate the time of motion:
Δy = v0yt + 0.5 ​ayt2
12 = 0*t + 0.5*9.8*t2
t = 1.56 sec
-        We will use the time to calculate the distance by substituting in the equation of horizontal direction:
ΔX = vxt
ΔX = 4*1.56 = 6.3 m