Acceleration is defined as the rate of change of velocity with time

**a = Δv/Δt**

Like displacement and velocity, acceleration is a vector quantity i.e it has direction as well as magnitude

- a = acceleration - in metres per (second)
^{2}(ms^{-2}) - Δv =
**change in velocity**- in metres per second (ms^{-1}) - Δt = the corresponding
**change in time**- in seconds (s)

Therefore **a = (v -u)/(t _{2}-t_{1})**

- a = acceleration - in metres per (second)
^{2}(ms^{-2}) - v =
**final velocity**- in metres per second (ms^{-1}) - u =
**initial velocity**- in metres per second (ms^{-1}) - t
_{2}= the time at which the final velocity occurs - in seconds (s) - t
_{1}= the time at which the initial velocity occurs - in seconds (s)

** Variations on how you may see the equation written**

We would normally begin timing when the object is just starting to accelerate from its initial velocity so t_{1}= 0

In this case we would just use **t** to represent the final time (rather than t_{2}).
So we would commonly see the equation written as

**a = (v -u)/(t)**

For examples where the initial velocity (u) is zero we would just write

**a = v/t**

Note in "everyday language" acceleration is often taken to mean increase in speed. It is important to note that in Physics
the correct definition of rate of change of velocity covers **increase in speed, decrease in speed** and **change in direction**
(remember velocity is a vector quantity it has a specific direction so if direction changes then velocity has changed).
(The term deceleration can still be used which specifically refers to a change in velocity caused by a decrease in speed.)

**Example 1**

The velocity of an object is 2ms^{-1} when the time is 2 seconds and increases to 14ms^{-1} when the time is 5 seconds.
What is it's acceleration?

u = 2ms^{-1}

v = 14ms^{-1}

t_{1} = 2s

t_{2} = 5s

a= Δv/Δt

a = (v - u)/(t_{2} - t_{1}) = 12/3 = **4ms ^{-2}**

**Example 2**

The velocity of an object increases from 4ms^{-1} to 16ms^{-1} in 24 seconds. What is it's acceleration?

u = 4ms^{-1}

v = 16ms^{-1}

t = 24s

a= Δv/Δt

a = (v - u)/(t) = 12/24= **0.5ms ^{-2}**

**Example 1**

The velocity of an initially stationary object increases to 3ms^{-1} in 0.5 seconds. What is its acceleration?

v = 3ms^{-1}

t = 0.5s

a= v/t

a = 3/0.5 =**6ms ^{-2}**