Tangential acceleration concept is applied to measure how the tangential velocity of a point at a certain radius changes with time. Tangential acceleration is similar to linear acceleration but specific to the tangential direction which relates to circular motion.In other words, the rate of change of tangential velocity of a particle in a circular orbit is known as Tangential acceleration. It directs towards tangent to the path of the body.

## Tangential Acceleration Formula

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

## Tangential Acceleration Formula In Terms Of Distance

\(a_{t}=\frac{\mathrm{d}^{2} s}{\mathrm{d} t^{2}}\)
Or \(a_{t}=v.\frac{dv}{ds}\) |

## Notations Used In The Formula

- a
_{t}is the tangential acceleration - Δv is the change in the angular velocity
- Δt is the change in time
- v is the linear velocity
- s is the distance covered
- t is teh time taken

Tangential acceleration formula is applied to calculate the tangential acceleration and the parameters related to it.It is expressed in **meter per sec square****(m/s ^{2})**.

### Linear Acceleration Formula

Linear acceleration is defined as the uniform acceleration caused by a moving body in a straight line. There are three equations that are important in linear acceleration depending upon the parameters like initial and final velocity, displacement, time and acceleration.

Following is the table explain all the three equations that are used in linear acceleration:

First equation of motion | v=u+at |

Second equation of motion | \(s=ut+\frac{1}{2}at^{2}\) |

The third equation of motion | \(v^{2}=u^{2}+2as\) |

### Notations Used In The Formula

- u is the initial velocity
- a is the acceleration
- t is the time taken
- v is the final velocity
- s is the acceleration

### Solved Examples

**Example 1: **A body accelerates uniformly on a circular path with a speed of 30 m/s to 70m/s in 20s. Calculate its tangential acceleration.

**Solution:**Given:

Initial velocity vi = 30 m/s, Final velocity v_{f} = 70 m/s

Change in velocity dv = v_{f} – v_{i} = 70 – 30 = 40 m/s

Time taken dt = t_{f} – t_{i} = 20 – 0 = 20s

The tangential acceleration is given by at = dv / dt = 40 / 20 = 2 m/s^{2}.

**Example2:**A runner starts from rest and accelerates at a uniform rate 10m/s in the time interval of 5s, running on a circular track of radius 50m. Determine the tangential acceleration.

**Solution:**Given: Initial velocity v_{i} = 0

Final velocity v_{f} = 10 m/s

Change in velocity dv = v_{f} – v_{i} = 10 – 0 = 10 m/s

Time taken dt = 5s

The tangential acceleration is given by at = dv / dt = 10 / 5 = 2m/s^{2}.

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