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 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 **

**Table -2**

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.**

Time(s) |
Velocity (m) |

1 | 2 |

2 | 4 |

3 | 6 |

4 | 8 |

5 | 10 |

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 could be mathematically defined as,

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;

The above equation is also known as**constant acceleration formula.**

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

where,

v

v

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

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 + $\frac{1}{2}$ 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

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

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 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,$\alpha$ = $\frac{\alpha_{T}}{r}$

$\omega$ = angular velocity of the object

$\alpha$ = angular acceleration of the object

$\alpha_{T}$ = linear tangential acceleration

$\theta$ = angular motion of the object

t = time

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,

Solving it further we have,

**v = v**_{0} - $\alpha$ t

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

where,

$\omega$ = Angular Velocity of the object

$\alpha$ = Angular Acceleration of the object

t = Time

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

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

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

Solving it further we have,

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

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$^{2}$.

Times(s) |
Velocity(m) |

1 |
2 |

2 |
4 |

3 |
6 |

4 |
8 |

5 |
10 |

Find the acceleration of the object?

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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