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Constant Acceleration

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.


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Constant Acceleration Definition

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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 table-1 and table-2, we can see that in the table-1 the velocity is constantly varying with same amount per second so the acceleration in this case is constant. Now if we look at table-2, 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.

Constant Acceleration Equations

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Acceleration could be mathematically defined as,

a = $\frac{v_{2}-v_{1}}{t_{2}-t_{1}}$

where,
v2 = velocity of the body at time t2
v1 = velocity of the body at time t1

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 = $\lim_{t \to 0}$ $\frac{\mathrm{d}v}{\mathrm{d}t}$

The above equation is also known as constant acceleration formula.

Constant Acceleration Graph

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Considering the above table 1, the graph of constant acceleration could be,

Constant Acceleration Graph

Motion with Constant Acceleration

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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/sec2With 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 + $\frac{1}{2}$ at2
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

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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 2-D and 3-D 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,
$\alpha$ = $\frac{\mathrm{d}\omega}{\mathrm{d}t}$ = $\frac{\mathrm{d}^{2}\theta}{\mathrm{d}t^{2}}$

$\alpha$ = $\frac{\alpha_{T}}{r}$
Where,
$\omega$ = angular velocity of the object
$\alpha$ = angular acceleration of the object
$\alpha_{T}$ = linear tangential acceleration
$\theta$ = angular motion of the object
t = time
→ Read More

Angular Motion with Constant Acceleration

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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 v0 at time t = 0 and velocity at time t is v, then the acceleration can be obtained as,

$\alpha$ = $\frac{v - v_{0}}{t-0}$

$\alpha$ = $\frac{v - v_{0}}{t}$

Solving it further we have,
v = v0 - $\alpha$ t

Considering that the body is moving in the angular direction, we can find the angular acceleration of the object also,

$\alpha$ = $\frac{\omega }{t}$

where,
$\omega$ = Angular Velocity of the object
$\alpha$ = Angular Acceleration of the object
t = Time

Constant Acceleration Problems

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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 + $\frac{1}{2}$at$^{2}$
where, u is the initial velocity.

Putting the values of all the variables from the question we have;
   a = $\frac{2s}{t^{2}}$
   a = $\frac{2\times100}{(5)^{2}}$
Solving we have
    a = 8 m/sec$^{2}$

So, the acceleration of the bus is 8 m/sec$^{2}$.
 

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$_{1 - 3}$ = $\frac{v_{3}-v_{1}}{t_{3}-t_{1}}$
a$_{1 - 3}$ = $\frac{6 - 2}{3 - 1}$
a$_{1 - 3}$ = $\frac{4}{2}$ = 2m/sec$^{2}$
a$_{2 - 5}$ = $\frac{v_{5} - v_{2}}{t_{5} - t_{2}}$
a$_{2 - 5}$ = $\frac{10 - 4}{5 - 2}$
a$_{2 - 5}$ = $\frac{6}{3}$ = 2m/sec$^{2}$

As we see that the a$_{1 - 3}$ and a$_{2 - 5}$ 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.
 

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