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.).
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
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