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There are many influencing parameters are present while motion of an object. The important parameters are distance, displacement, velocity, acceleration etc. Here we are highlighting the concept of constant acceleration. This is same as that of acceleration but the only difference is it will accelerate at a constant rate. Mainly there are three equation of motion is present if the acceleration is constant in both magnitude and direction. If the motion of an object with the constant acceleration we can say that it follows a fixed path. Each equation involves the parameters mentioned above.The nature of this physical quantity is described below in detail.

Constant acceleration is a kind of acceleration, in which a body changes its velocity by equal amount per second. If the net force acting on an object is constant, then the object's acceleration is also constant. If the net force is zero, the acceleration is zero and the object's velocity does not change.
There are three simple equations may be used to represent motion of an object when it is traveled with a constant acceleration. If an object moving with an initial velocity u and it is moving in a straight line with a constant acceleration a, the following equations represent the motion of that body.v = u + at
s = ut + $\frac{1}{2}$ at^{2}
$v^{2}$ = $u^{2}+2as$
Where v is the final velocitys = ut + $\frac{1}{2}$ at^{2}
$v^{2}$ = $u^{2}+2as$
u is the initial velocity
a is the constant acceleration
t is the time taken
s is the displacement
These equations cannot be used if the acceleration is not constant.
The given graph shows the relation between the velocity and time in the positive direction. The slope will give the constant acceleration.
According to Newton's Universal law of gravitation, the gravitational force between any two bodies is directly proportional to the product of magnitudes of the masses and inversely proportional to the square of the distance between them. The mathematical representation of this law is given as,
Where F is the gravitational force
G is the gravitational acceleration constant, which is given by G = 6.67×10^{11} Nm^{2}/Kg^{2} .
M is the mass of a larger body
m is the mass of the lighter body
r is the distance between two bodies
F= G $\frac{Mm}{r^{2}}$
Where F is the gravitational force
G is the gravitational acceleration constant, which is given by G = 6.67×10^{11} Nm^{2}/Kg^{2} .
M is the mass of a larger body
m is the mass of the lighter body
r is the distance between two bodies
^{ }
Consider an object dropped from particular height, it get accelerated due to the gravity. The value of acceleration due to gravity is given as 9.8 m/s^{2}^{ }. There are three important relations present in a motion with constant acceleration. One equation relates the displacement with constant acceleration. Using this equation we can find out the displacement of an object moving with a constant acceleration. The second equation gives an idea about the relation between velocity of an object with constant acceleration. So, we can calculate the velocity of an object with the constant acceleration. The third equation connects the velocity and displacement with constant acceleration. These three important equations are mentioned above in the constant acceleration equations section. The more details about each equations are given in following sections.
Let us assume a situation in which the velocity of an object is continuously changing but the acceleration is constant. If the initial velocity is taken as v_{i} at time t_{0} and the the final velocity is v at time t. So the angular acceleration represented by an equation, which is equal to
$\alpha$ = $\frac{vv_{i}}{t0}$
$\alpha$ = $\frac{vv_{i}}{t}$
Rearranging this equation we get,
v = $v_{i} + \alpha{t}$
When the body is moving with an angular velocity, the equation for the angular acceleration is also calculated as
Where α is the angular acceleration
ω is the angular velocity
t is the time taken
$\alpha$ = $\frac{vv_{i}}{t0}$
$\alpha$ = $\frac{vv_{i}}{t}$
Rearranging this equation we get,
v = $v_{i} + \alpha{t}$
When the body is moving with an angular velocity, the equation for the angular acceleration is also calculated as
$\alpha$ = $\frac{\omega }{t}$
Where α is the angular acceleration
ω is the angular velocity
t is the time taken
One of the important equations in constant acceleration is the relation connecting between the displacement and constant acceleration. The other parameters in this equation are initial velocity and the time. The mathematical representation of this equation is given by,
u is the initial velocity of an object
a is the constant acceleration
t is the time taken to an object to cover the displacement s
Using this equation one can find the displacement of an object moving with constant acceleration.
s = ut + $\frac{1}{2}at^{2}$
Where s is the displacement of an objectu is the initial velocity of an object
a is the constant acceleration
t is the time taken to an object to cover the displacement s
Using this equation one can find the displacement of an object moving with constant acceleration.
The relation connecting between the velocity and constant acceleration is given by,
u is the initial velocity
t is the time taken
Another equation which connects the velocity and displacement with the constant acceleration is
u is the initial velocity
a is the constant acceleration
s is the displacement
t is the time taken
v = u + at
Here v is the final velocityu is the initial velocity
t is the time taken
Another equation which connects the velocity and displacement with the constant acceleration is
$v^{2}$ = $ u^{2}+2as $
Where v is the final velocityu is the initial velocity
a is the constant acceleration
s is the displacement
t is the time taken
Non constant acceleration describes the general motion of an object. It is given as the rate of change of velocity. That is acceleration is varying during the motion of an object. This changes can be described in terms of time (t) or position (x). We can also represent the constant acceleration in terms of velocity. It is clear that, if we can express the non constant acceleration in one dimension, we can also extend it in to two and three dimension. So, now we can see the equations of non constant acceleration in one dimension.
Consider the equation of acceleration in x, given as the function a(x). So, the equation is given by,
a(x) = $\frac{dv}{dt}$
Rearrange the given equation we get,
a(x) = $\frac{dv}{dx}$ $\times$ $\frac{dx}{dt} $ = v $\frac{dv}{dx}$
By rearranging and integrating the above equation :
$\int vdv$ = $a(x) dx$
This relation gives a relation of velocity in terms x. We can find out the expression for position or displacement by using this equation, on rearranging and integrating :
v(x) = $\frac{dx}{dt}$
dt = $\frac{dx}{v(x)}$
$\Delta t$ = $\int \frac{dx}{v(x)}$
This equation gives the relation between the position and t.
Consider the equation of acceleration in x, given as the function a(t). So, the equation is given by,
a(t)= $\frac{dv}{dt}$
Rearrange the given equation we get,
dv = a(t)dt
$\Delta v$ = $\int a(t)dt$
This relation gives a relation of velocity in terms t. We can find out the expression for position or displacement by using this equation, on rearranging and integrating :
v(t) = $ \frac{dx}{dt} $
dx = v(t)dt
$\Delta x$ = $\int v(t)dt$
This equation gives an expression of position in t after using initial conditions.
a(x) = $\frac{dv}{dt}$
Rearrange the given equation we get,
a(x) = $\frac{dv}{dx}$ $\times$ $\frac{dx}{dt} $ = v $\frac{dv}{dx}$
By rearranging and integrating the above equation :
$\int vdv$ = $a(x) dx$
This relation gives a relation of velocity in terms x. We can find out the expression for position or displacement by using this equation, on rearranging and integrating :
v(x) = $\frac{dx}{dt}$
dt = $\frac{dx}{v(x)}$
$\Delta t$ = $\int \frac{dx}{v(x)}$
This equation gives the relation between the position and t.
Consider the equation of acceleration in x, given as the function a(t). So, the equation is given by,
a(t)= $\frac{dv}{dt}$
Rearrange the given equation we get,
dv = a(t)dt
$\Delta v$ = $\int a(t)dt$
This relation gives a relation of velocity in terms t. We can find out the expression for position or displacement by using this equation, on rearranging and integrating :
v(t) = $ \frac{dx}{dt} $
dx = v(t)dt
$\Delta x$ = $\int v(t)dt$
This equation gives an expression of position in t after using initial conditions.
 Motion of a freely falling body
 Motion of a plane
 Motion of a spherical body on an inclined plane if there no unbalanced force is acting on it
 Projectile motion
Solved Examples
Question 1: A bike traveling at a speed of 10 m/s and it increases its speed to 30 m/s with a constant acceleration of 1.25 m/s^{2} . What is the time taken and how far it can move with this time?
Solution:
The given parameters are initial velocity u = 10 m/s, final velocity v = 30 m/s and constant acceleration a = 1.25 m/s^{2}
We know the equation of motion with constant acceleration is
v = u + at
That is, 30 = 10 + 1.25t
t = $\frac{20}{1.25}$
t = 16 s
To find out the displacement, we have the equation
s = ut+ $\frac{1}{2}$ $at^{2}$
s = 10 $ \times 16 $ + $ \frac{1}{2}\times 1.25\times 16^{2} $
s = 320 m
Solution:
The given parameters are initial velocity u = 10 m/s, final velocity v = 30 m/s and constant acceleration a = 1.25 m/s^{2}
We know the equation of motion with constant acceleration is
v = u + at
That is, 30 = 10 + 1.25t
t = $\frac{20}{1.25}$
t = 16 s
To find out the displacement, we have the equation
s = ut+ $\frac{1}{2}$ $at^{2}$
s = 10 $ \times 16 $ + $ \frac{1}{2}\times 1.25\times 16^{2} $
s = 320 m
Question 2: A boy throws a ball vertically upward with a speed of 7.7 m/s from the ground. If there is no air resistance is acting on this ball it moves under the gravity. Calculate the maximum height at which this ball can attain?
Solution:
It is given that, u = 7.7m/s, a = 9.81 m/s^{2} (negative sign because the motion of the ball is in upward direction), and from the question it is clear that at maximum height v = 0 m/s
The expression of motion with constant acceleration is given by
$v^{2} = u^{2}+2as$
$0 = 7.7^{2} + 2 \times 9.81 \times s $
s = 3 m
Solution:
It is given that, u = 7.7m/s, a = 9.81 m/s^{2} (negative sign because the motion of the ball is in upward direction), and from the question it is clear that at maximum height v = 0 m/s
The expression of motion with constant acceleration is given by
$v^{2} = u^{2}+2as$
$0 = 7.7^{2} + 2 \times 9.81 \times s $
s = 3 m