When an object is moving then there is change in its speed and also its velocity. The change in speed is called velocity while the change in velocity is known as acceleration. If an object is said to be in acceleration then its velocity is continuously changed. This is the state of acceleration. This is defined as the rate change of velocity with time.
This word acceleration is generally used to represent the increasing speed state. It is not necessary to increase the speed, it also depends on velocity change and change in the direction of motion with time. This is the reason that it is a vector quantity which has both the direction and magnitude. Thus it can be occurred with increasing or decreasing or changing the direction of object motion.
Here we discuss about the different types of acceleration especially constant acceleration in which the velocity of moving object does not change in a particular period of time. Now we discuss its formula, motion graph, and some problem based on it.
Constant acceleration is the special case of acceleration. When a moving body changes its velocity by the equal amount per second, the body is said to be moving with the constant accelerationFor better understanding of the constant acceleration, consider the following tables:
Table  1
Time(s) 
Velocity (m) 
1 
2 
2 
4 
3 
6 
4 
8 
5 
10

Table 2
Time(s) 
Velocity (m) 
1 
0 
2 
3 
3 
4 
4 
7 
5 
12

If we look at table1 and table2, we can see that in the table1 the velocity is constantly varying with same amount per second so the
acceleration in this case is constant. Now if we look at table2, We can conclude that the velocity in this case is not varying constantly per second and hence the
acceleration in this case is not constant.
From the above table it is concluded that the body moving with constant acceleration should have constant velocity per unit time.
Acceleration could be mathematically defined as,
a = v2−v1t2−t1$\frac{{v}_{2}{v}_{1}}{{t}_{2}{t}_{1}}$
where,
v
_{2} = velocity of the body at time t
_{2}
v
_{1} = velocity of the body at time t
_{1}
For the body to move with constant acceleration the difference between its velocities between equal time intervals should be equal.
If the time interval is considered to be very small then the above equation could be rewritten as;
a = limt→0$\underset{t\to 0}{lim}$ dvdt$\frac{\mathrm{d}v}{\mathrm{d}t}$
The above equation is also known as
constant acceleration formula.
Considering the above table 1, the
graph of constant acceleration could be,
When a body is dropped from a height, it gets accelerated under the influence of gravity and its acceleration is equal to 9.8 m/sec^{2}. With the help of this statement, we can find the height from which the body is dropped
. The acceleration in this case is constant. The relation between acceleration and displacement of the body is,
s = ut + 12$\frac{1}{2}$ at^{2}
Where,
s = displacement of the object
u = initial velocity of the object
a = acceleration (or deceleration) of the object
t = time taken by the object to get displaced by ‘s’ units
Non constant acceleration is the most general description of motion. It is the rate of change in velocity. In other words, it means that acceleration changes during motion of the object. This variation can be expressed either in terms of position (x) or time (t).If the non constant acceleration is described in one dimension, we can easily extend the analysis to two or three dimensions using composition of motions in component directions.
For this reason, we shall confine ourselves to the consideration of non constant, i.e., variable acceleration in one dimension. For finding the 2D and 3D equations of the angular motion, we would find the equation of motion in each direction separately since the motion in one direction is independent of the motion in other direction.
Angular Acceleration is the rate of change of the angular velocity of the object with respect to time.
Angular acceleration can also be defined, mathematically, as shown below,
α$\alpha $ = dωdt$\frac{\mathrm{d}\omega}{\mathrm{d}t}$ = d2θdt2$\frac{{\mathrm{d}}^{2}\theta}{\mathrm{d}{t}^{2}}$
α$\alpha $ = αTr$\frac{{\alpha}_{T}}{r}$
Where,
ω$\omega $ = angular velocity of the object
α$\alpha $ = angular acceleration of the object
αT${\alpha}_{T}$ = linear tangential acceleration
θ$\theta $ = angular motion of the object
t = time
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Angular Motion with Constant Acceleration
Consider the situation that an object is moving in a plain such that it is changing its direction frequently. The acceleration of the object is, say constant. Lets also consider that the initial velocity v
_{0} at time t = 0 and velocity at time t is v, then the acceleration can be obtained as,
α$\alpha $ = v−v0t−0$\frac{v{v}_{0}}{t0}$
α$\alpha $ = v−v0t$\frac{v{v}_{0}}{t}$
Solving it further we have,
v = v_{0}  α$\alpha $ t
Considering that the body is moving in the angular direction, we can find the angular acceleration of the object also,
α$\alpha $ = ωt$\frac{\omega}{t}$
where,
ω$\omega $ = Angular Velocity of the object
α$\alpha $ = Angular Acceleration of the object
t = Time
Constant Acceleration Problems
Lets take up some problems to understand the uniform or constant acceleration better.
Solved Examples
Question 1: A bus starts from rest and accelerates uniformly over a time of 5 sec. In this time it covered a distance of 100 m. Find the acceleration of the bus?
Solution:
From question, we have;
s = 100m
t = 5 sec
Now using following equation;
s = u×$\times $t + 12$\frac{1}{2}$at2${}^{2}$
where, u is the initial velocity.
Putting the values of all the variables from the question we have;
a = 2st2$\frac{2s}{{t}^{2}}$
a = 2×100(5)2$\frac{2\times 100}{(5{)}^{2}}$
Solving we have
a = 8 m/sec2${}^{2}$
So, the acceleration of the bus is 8 m/sec2${}^{2}$.
Question 2: For the better understanding of the constant acceleration, let us consider following table:
Times(s) 
Velocity(m) 
1 
2 
2 
4 
3 
6 
4 
8 
5 
10 
Find the acceleration of the object?
Solution:
For checking whether the object is traveling with constant velocity, find the acceleration between any two points and checks it with another two points. If both have equal results then it could be concluded that the object is moving with constant acceleration.
So, we will consider motion of the object from 1 sec to 3 sec and 2 sec to 5 sec.
a
1−3${}_{13}$ =
v3−v1t3−t1$\frac{{v}_{3}{v}_{1}}{{t}_{3}{t}_{1}}$
a
1−3${}_{13}$ =
6−23−1$\frac{62}{31}$
a
1−3${}_{13}$ =
42$\frac{4}{2}$ = 2m/sec
2${}^{2}$
a
2−5${}_{25}$ =
v5−v2t5−t2$\frac{{v}_{5}{v}_{2}}{{t}_{5}{t}_{2}}$
a
2−5${}_{25}$ =
10−45−2$\frac{104}{52}$
a
2−5${}_{25}$ =
63$\frac{6}{3}$ = 2m/sec
2${}^{2}$
As we see that the a
1−3${}_{13}$ and a
2−5${}_{25}$ are equal we can consider that the body is moving with constant acceleration.
This can also be obtained from the graph between the velocity and time.
When an object is in motion then its velocity is changed continuously then it is called in acceleration state. We know that acceleration is the change in velocity of moving object with respect to time. When an object is moved on circular path then its velocity is the angular velocity. The angular velocity is in the perpendicular direction of the rotational plane and it is related to the change in the angular speed and measure with unit as radians per second or revolutions per second.
The angular acceleration shows the change in the angular velocity with time. It is not required to be this rotational acceleration in the same direction of angular velocity; for example, if a car is rolling on highway with increasing its speed then the direction of angular acceleration is on the left side of the axis of the wheel and it becomes disappears when the car stops and maintained its constant velocity. When it slows down then the acceleration is in the reverse direction. It also helps to give relation between circular motion and curve motion.
Here, we are discussing about the angular acceleration and its mathematical formula, and relation with torque.
Angular Acceleration Definitio
We know that the rate of change of displacement with respect to time is called velocity of the object. But if there is a change in the velocity while the body is going in uniform circular motion, what can we call it? how can we Calculate it?
Angular acceleration is the rate of change of angular Velocity with respect to time. It is a vector quantity.
It can be represented as:
α$\alpha $ = dωdt$\frac{\mathrm{d}\omega}{\mathrm{d}t}$ = d2θdt2$\frac{{\mathrm{d}}^{2}\theta}{\mathrm{d}{t}^{2}}$.
or
α$\alpha $ = atr$\frac{{a}_{t}}{r}$
where,
ω$\omega $ is the angular Velocity
a_{t} is the linear tangential acceleration
r is the radius of Circular path
If the angular velocity is constant, then algebraically it is defined as,
ω$\omega $ = θt$\frac{\theta}{t}$
where,
ω$\omega $ is angular velocity
θ$\theta $ is the angle rotated
‘t’ is the time taken
It may be realized when the angular velocity is constant, the angular acceleration is 0. If the velocity is not constant, then the constant angular acceleration α$\alpha $ is defined as
α$\alpha $ = ωt$\frac{\omega}{t}$ = θt2$\frac{\theta}{{t}^{2}}$
If the angular acceleration is not constant and varies from time to time, then we can only refer to average angular acceleration and instantaneous acceleration.
In such a case the angular acceleration formulas are,
α$\alpha $_{av} = (ω2–ω1)(t2–t1)$\frac{({\omega}_{2}\u2013{\omega}_{1})}{({t}_{2}\u2013{t}_{1})}$ and α$\alpha $_{i} = dωdt$\frac{\mathrm{d}\omega}{dt}$ = d2θdt2$\frac{{\mathrm{d}}^{2}\theta}{\mathrm{d}{t}^{2}}$.These equations help us in finding angular acceleration.
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Angular Acceleration Units
As per the definition, an angular acceleration is the amount of angle covered per square of time. Hence dimensionally its unit must be the ratio of angle to the square of time. Then, in common terms, the angular acceleration unit is degrees/square of time like s^{2}, min^{2}, hr^{2}. But in internationally accepted unit (SI unit), the unit of angle is radian (rad) and the unit of time is second (s). Hence the standard unit of Angular acceleration is rad/s^{2}.
Hence it is more sensible to define the displacement in terms of the angle it has rotated. If the object completes one full rotation it comes back to the initial position on the circle, meaning the displacement is 0. But the angular displacement is 360^{o} degrees or 2π$\pi $ radians. Therefore, in Circular motions, the rate of change of angle is defined as angular velocity and the rate of change of angular velocity is defined as ‘Angular acceleration’.
Average Angular Acceleration
Angular acceleration is the rate of change of angular velocity with respect to time. Suppose the body has moved some distance then if we consider initial and final point where the displacement gets an end, then its average angular acceleration is given by
Average angular acceleration,
αav=ωf−ωitf−ti${\alpha}_{av}=\frac{{\omega}_{f}{\omega}_{i}}{{t}_{f}{t}_{i}}$
where,
ω_{f }= final angular velocity
ω_{i} = initial angular velocity
t_{f} = final interval of time or the end of time
t_{i }= initial interval of time
It has the same unit as the angular acceleration, i.e., rad/s2.
Torque and Angular Acceleration
In case of linear motion, as per Newton’s second law a force is required to accelerate object. That force is defined as the product of the mass of the object and the acceleration created.
In case of circular motion the force that is required to impart angular acceleration is called ‘Torque’. In other words, torque is an angular force and it is denoted by the Greek letter τ$\tau $ (pronounced as ‘tau’).
Also in rotational motion the moment of inertia I of the object plays the role of mass.
The torque in a circular motion is defined as,
τ$\tau $ = I α$\alpha $
Constant Angular acceleration
If an object undergoes rotational motion about a fixed axis under a constant angular acceleration α$\alpha $, its motion can be described with the following set of equations,
ω$\omega $  ω0${\omega}_{0}$ = α$\alpha $ t.
θ$\theta $  θ0${\theta}_{0}$ = ω0${\omega}_{0}$ t + 12$\frac{1}{2}$ α$\alpha $ t^{2}
ω2${\omega}^{2}$ = ω20${\omega}_{0}^{2}$ + 2 α$\alpha $ (θ$\theta $  θ0${\theta}_{0}$)
where ω0${\omega}_{0}$ = angular speed of the rigid body at time t=0
ω$\omega $ = angular speed of the rigid body at time t
α$\alpha $ = angular accelerationAngular Acceleration to Linear Acceleration
As mentioned earlier, the definition of average angular velocity is the change in angle θ$\theta $ with respect to time t.
ω$\omega $ = θt$\frac{\theta}{t}$
where,
θ$\theta $ is the angle rotated in the time t
Now we will bring in the linear velocity v and the radius of the circle r, in a circular motion.
Let ι$\iota $ be the actual distance moved by the object along the circumference. As per the geometry of circles,
θ$\theta $ = ιr$\frac{\iota}{r}$,
Therefore, ω$\omega $ = θt$\frac{\theta}{t}$
= ιrt$\frac{\iota}{rt}$ = ιt1r$\frac{\iota}{t}\frac{1}{r}$.
Now if we say, v = ιt$\frac{\iota}{t}$
we can write
ω$\omega $ = vr$\frac{v}{r}$ and hence,
α$\alpha $ = ωt$\frac{\omega}{t}$
where v is the linear velocity of the object at any point and its direction is along the tangent at that point.
Now considering an infinitesimal study,
ωi${\omega}_{i}$ = dθdt$\frac{\mathrm{d}\theta}{dt}$ = dιdt1r$\frac{\mathrm{d}\iota}{\mathrm{d}t}\frac{1}{r}$ =