Acceleration is defined as

The rate of change of velocity with respect to time.

Acceleration is a vector quantity as it has both magnitude as well as direction. It is also the second derivative of position with respect to time or it is a first derivative of velocity with respect to time.

Instantaneous acceleration is defined as

The ratio of change in velocity during a given time interval such that the time interval goes to zero.

Acceleration formula is given as:

\(acceleration = \frac{(final\;velocity)-(initial\;velocity)}{time}\)

\(acceleration = \frac{change\;in\;velocity}{time}\)

\(a= \frac{v_{f}-v_{i}}{t}\)

\(a= \frac{\Delta v}{t}\) |

Where,

- a is the acceleration in m.s
^{-2} - v
_{f }is the final velocity in m.s^{-1} - v
_{i}is the initial velocity in m.s^{-1} - t is the time interval in s
- Δv is the small change in the velocity in m.s
^{-1}

The SI unit of acceleration is given as:

SI unit | m/s^{2} |

So can we have a situation when speed remains constant but the body is accelerated? Actually, it is possible in circular where speed remains constant but since the direction is changing hence the velocity changes and the body is said to be accelerated.

The average acceleration over a period of time is defined as the total change in velocity in the given interval divided by the total time taken for the change. For a given interval of time, it is denoted as *ā*.

Mathematically,

Where *v _{2}* and

**Average acceleration:** In the velocity-time graph shown above, the slope of the line between the time interval *t _{1}* and

**Instantaneous acceleration: **In a velocity-time curve, the instantaneous acceleration is given by the slope of the tangent on the v-t curve at any instant.

Consider the velocity-time graph shown above. Here, between the time intervals of 0-2 seconds, the velocity of the particle is increasing with respect to time; hence the body is experiencing a positive acceleration as the slope of the v-t curve in this time interval is positive.

Between the time intervals of 2-3 seconds, the velocity of the object is constant with respect to time; hence the body is experiencing zero acceleration as the slope of the v-t curve in this time interval is 0.

Now, between the time intervals of 3-5 seconds, the velocity of the body is decreasing with respect to time; hence the body experiences a negative value of the rate of change of velocity as the slope of the v-t curve in this time interval is negative.

**Q1. What will be the acceleration of an object which moves with uniform velocity?**

**Ans:** Given,

The velocity is uniform, therefore the initial and final velocities are equal and is given as V.

∴From definition, acceleration is given as:

\(a= \frac{v_{f}-v_{i}}{t}\) \(a=\frac{0}{t}\) ∴ a = 0

**Q2. A truck is moving with a constant velocity, v = 5 m.s ^{-1}. The driver stops for diesel and the truck accelerates forward. After 20 seconds, the driver stops accelerating to maintain a constant velocity, v = 25 m.s^{-1}. What is the truck’s acceleration?**

Initial velocity, v

Final velocity, v

Time interval, t = 20 s

∴From definition, acceleration is given as:

\(a= \frac{v_{f}-v_{i}}{t}\) \(a=\frac{25-5}{20}\) ∴a=1 m.s

**Q3. Find the final velocity of the ball that is dropped from the second floor if the ball takes 18 seconds before hitting the ground. The acceleration due to gravity is g = 9.80 m.s ^{-2}.**

Initial velocity, v

Final velocity, v

Acceleration due to gravity = a = g = 9.80 m.s

Time interval, t = 18 s

∴From definition, acceleration is given as:

\(a= \frac{v_{f}-v_{i}}{t}\) Rearranging the formula,

v

v

v_{f}=176.4 m.s^{-1}

Following is the table of acceleration vs velocity:

Parameter |
Acceleration |
Velocity |

Definition | Acceleration is defined as the change in the velocity of an object with respect to time | Velocity is defined as the speed of an object in a particular direction |

Formula | Velocity/Time | Displacement/Time |

Unit | m.s^{-2} |
m.s^{-1} |