You might have observed while pushing a bus it suddenly starts. Lift gives upward push when it starts. This is what acceleration is! Here Velocity changes. Hence the body accelerates. The rate of change of velocity of an object is known as Acceleration. An object's acceleration is the net result of any and all forces acting on the object, as described by Newton's Second Law. Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity.

**Acceleration is the rate of change in velocity to the change in time. It is represented by symbol a and is given by **

The **S.I** unit for Acceleration is meter per second squared or m/s^{2}.

If initial velocity u, final velocity v time taken t are given. Then the acceleration is given by formula

Where,

v = Final Velocity,

u = Initial velocity,

a = acceleration,

t = time taken,

s = distance traveled.

If initial velocity u, final velocity v time taken t are given. Then the acceleration is given by formula

Where,

v = Final Velocity,

u = Initial velocity,

a = acceleration,

t = time taken,

s = distance traveled.

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Acceleration Formula Calculator | Acceleration due to Gravity Calculator |

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Below are given problems based on acceleration which helps you to get an idea how the formula is used.

### Solved Examples

**Question 1: **A toy car accelerates from 3m/s to 5m/s in 5 s. What is its acceleration?

** Solution: **

Given: Initial Velocity u = 3m/s,

Final Velocity v = 5m/s,

Time taken t = 5s.

The Acceleration is given by a = $\frac{v - u}{t}$

= $\frac{5 - 3}{5}$

= $\frac{2}{5}$

= 0.4 m/s^{2}.

**Question 2: **A stone is dropped into the river from the bridge. The Stone takes 4s to touch the water surface of the river. Calculate the height of the bridge from the water level.

** Solution: **

given: Initial Velocity u = 0 ( as the stone was at rest),

Time taken t = 5s,

Acceleration due to gravity is a = g = 9.8 m/s^{2},

Height of bridge = distance traveled by stone = s

The distance traveled is given by

s = ut + $\frac{1}{2}$ gt^{2}

= 0 + $\frac{1}{2}$ $\times$ 9.8 m/s^{2} $\times$ (5s)^{2}

= 122.5 m.

**Question 3: **John throws a ball up with a velocity of 15 m/s. what is its acceleration if travels a distance of 11.4 m?

** Solution: **

Given: Initial velocity u = 15m/s,

Final velocity v = 0,

Distance traveled s = 11.4 m,

To find the acceleration use the formula:

v^{2} = u^{2} + 2as

a = $\frac{v^{2} - u^{2}}{2g}$

= $\frac{0 - 15^{2}}{2 \times 11.4}$

= $\frac{- 225}{22.8}$

= -9.8 m/s^{2}.

The retardation is taking place at the rate of gravity.

Given: Initial Velocity u = 3m/s,

Final Velocity v = 5m/s,

Time taken t = 5s.

The Acceleration is given by a = $\frac{v - u}{t}$

= $\frac{5 - 3}{5}$

= $\frac{2}{5}$

= 0.4 m/s

given: Initial Velocity u = 0 ( as the stone was at rest),

Time taken t = 5s,

Acceleration due to gravity is a = g = 9.8 m/s

Height of bridge = distance traveled by stone = s

The distance traveled is given by

s = ut + $\frac{1}{2}$ gt

= 0 + $\frac{1}{2}$ $\times$ 9.8 m/s

= 122.5 m.

Given: Initial velocity u = 15m/s,

Final velocity v = 0,

Distance traveled s = 11.4 m,

To find the acceleration use the formula:

v

a = $\frac{v^{2} - u^{2}}{2g}$

= $\frac{0 - 15^{2}}{2 \times 11.4}$

= $\frac{- 225}{22.8}$

= -9.8 m/s

The retardation is taking place at the rate of gravity.

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Centripetal Acceleration Formula | Angular Acceleration Formula |

Average Acceleration Formula | Angular Momentum Formula |

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