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# Average Acceleration: Definition, Formula, Examples and more

In this article, you will learn what we mean by average acceleration when describing the motion of a particle.

We will see the definition and formula for average acceleration as well as examples that show how to use the formula in practice.

We will also discuss other important things like how to find the average acceleration from a velocity vs time graph.

## Definition and formula for average acceleration

Let's consider a particle that moves along a straight line:

If at an instant t1 the particle has a velocity v1 and at a subsequent instant t2 it has a velocity v2, we indicate the change in velocity between t1 and t2, i.e., v2 β v1, with Ξv (delta-v) :

v2 β v1 = Ξv

Also, we indicate the interval of time between t1 and t2, i.e., t2 β t1, with Ξt (delta-t) :

t2 β t1 = Ξt

The average acceleration a12 that the particle has between instant t1 and instant t2 is defined as the ratio of the change in velocity between t1 and t2, i.e., v2 β v1, to the interval of time between t1 and t2, i.e., t2 β t1.

Therefore, the formula for average acceleration is

 a12 = v2 β v1 = Ξv t2 β t1 Ξt

In other words, the average acceleration a for an interval of time Ξt is equal to the change in velocity Ξv that occurs during that interval of time divided by the interval of time itself:

 a = Ξv Ξt

Thus, average acceleration is the average rate of change of velocity.

## Unit of average acceleration

Since average acceleration is the ratio of velocity to time, the SI unit of average acceleration is meters per second per second ((m/s)/s) or more simply meters per second squared (m/s2).

## Average velocity vs average acceleration

The difference between average velocity and average acceleration is that average velocity is the average rate of change of position, whereas average acceleration is the average rate of change of velocity:

 v12 = x2 β x1 = Ξx t2 β t1 Ξt
 a12 = v2 β v1 = Ξv t2 β t1 Ξt

## Example: Average acceleration of a car

A car goes from 0 to 100 km/h in 4.75 s. What is the average acceleration of the car?

Let's assume that the instant at which the car starts accelerating is 0 and let's label it as t1. So, the velocity v1 at instant t1 is 0.

t1 = 0, v1 = 0

We know that at a subsequent instant t2, the car reaches a velocity of 100 km/h. So, the velocity v2 at instant t2 is 100 km/h.

Also, since the time that passes between t1 and t2 is 4.75 s, and t1 is zero, it follows that t2 is 4.75 s.

t2 = 4.75 s, v2 = 100 km/h

We want to find the average acceleration between t1 and t2, which we label as a12.

The average acceleration a12 between an instant t1 and an instant t2 is equal to the ratio of the change in velocity that occurs between t1 and t2 to the interval of time between t1 and t2:

 a12 = v2 β v1 = 100 km/h β 0 = 100 km/h t2 β t1 4.75 s β 0 4.75 s

Before we can complete the calculation, we need to convert 100 km/h to m/s.

To convert from km/h to m/s, we divide the number by 3.6.

So, 100 km/h expressed in m/s is:

 100 m/s = 27.8 m/s 3.6

We can now complete the calculation of the average acceleration:

 a12 = 27.8 m/s = 5.85 m/s2 4.75 s

Thus, the average acceleration of the car is 5.85 m/s2.

## Example: Average acceleration of a train

A train, initially still, reaches a velocity of 40 km/h in 13 s. Then, it further increments its velocity by 35 km/h in 25 s. Calculate the average acceleration of the train for the first 13 s, the average acceleration for the following 25 s, and the overall average acceleration.

Let's suppose that the instant at which the train starts accelerating is 0 and label it as t1. So, the velocity v1 at instant t1 is 0.

t1 = 0, v1 = 0

At a subsequent instant t2, the train reaches a velocity of 40 km/h. So, the velocity v2 at instant t2 is 40 km/h.

The time that passes between the instant t1 and the instant t2 is 13 s, and since t1 is zero, it follows that t2 is 13 s.

t2 = 13 s, v2 = 40 km/h

At an instant t3, 25 s after t2, the train has incremented its velocity by 35 km/h. So, at instant t3, the train has a velocity v3 which is greater than v2 by 35 km/h.

v3 β v2 = 35 km/h
t3 β t2 = 25 s

Our task is to calculate the average acceleration of the train for the first 13 s, which start at t1 and end at t2, the following 25 s, which start at t2 and end at t3, and the overall average acceleration, i.e., the average acceleration from the starting instant t1 to the final instant t3.

Let's get started by calculating the average acceleration a12 between t1 and t2:

 a12 = v2 β v1 = 40 km/h β 0 = 40 km/h t2 β t1 13 s β 0 13 s
 40 km/h = 40 m/s = 11 m/s 3.6
 a12 = 11 m/s = 0.85 m/s2 13 s

Next, let's calculate the average acceleration a23 between t2 and t3:

 a23 = v3 β v2 = 35 km/h t3 β t2 25 s
 35 km/h = 35 m/s = 9.7 m/s 3.6
 a23 = 9.7 m/s = 0.39 m/s2 25 s

Last but not least, let's find the average acceleration a13 between t1 and t3:

 a13 = v3 β v1 t3 β t1

How can we determine what v3 β v1 is?

If we take a look at the v-axis, which shows the velocity of the train at different instants, we can see that the change in velocity v3 β v1 is equal to the sum of v2 β v1 and v3 β v2:

v3 β v1 = (v2 β v1) + (v3 β v2)
v3 β v1 = 40 km/h + 35 km/h
v3 β v1 = 75 km/h

With that in mind, it shouldn't be difficult to see that the interval of time t3 β t1 is equal to t2 β t1 and t3 β t2:

t3 β t1 = (t2 β t1) + (t3 β t2)
t3 β t1 = 13 s + 25 s
t3 β t1 = 38 s

Thus, the average acceleration a13 is

 a13 = 75 km/h 38 s
 75 km/h = 75 m/s = 21 m/s 3.6
 a13 = 21 m/s = 0.55 m/s2 38 s

Note that although in some cases the overall average acceleration a13 might be equal to the arithmetic mean of the average acceleration a12 and the average acceleration a23, you should not make that assumption when solving a problem.

Indeed, this is not the case for the train, as we can see by calculating the arithmetic mean of a12 and a23:

 a12 + a23 = 0.85 m/s2 + 0.39 m/s2 = 0.62 m/s2 2 2

Once again, do not assume that the average acceleration for the whole is equal to the arithmetic mean of the average accelerations for the parts.

## Sign of average acceleration

Since average acceleration is equal to the ratio of the change in velocity to the interval of time

 a = Ξv Ξt

and the interval of time Ξt is always positive, it follows that the sign of the average acceleration a is the same as the sign of the change in velocity Ξv.

The change in velocity Ξv can be positive, negative, or zero, depending on whether the particle incremented its velocity, decremented its velocity, or ended up with the same velocity with which it started.

Therefore, the average acceleration a can be positive, negative, or zero:

1. Ξv > 0 β a > 0
2. Ξv < 0 β a < 0
3. Ξv = 0 β a = 0

## Example: Average acceleration of an elevator

An elevator accelerates from rest for 9.35 s until it reaches a velocity of 7.83 m/s. Then, it continues moving with constant velocity. Finally, when the elevator is about to arrive at the destination, it decelerates for 12.5 s until it stops. Find the average acceleration of the elevator for the acceleration phase, for the deceleration phase, and from the moment it starts accelerating to the moment it decelerates to a stop.

Let's consider the instant at which the elevator starts accelerating to be 0 and refer to it as t1. Thus, the velocity v1 at instant t1 is 0.

t1 = 0, v1 = 0

At a subsequent instant t2, 9.35 s after t1, the elevator reaches a velocity of 7.83 m/s. Therefore, the velocity v2 at instant t2 is equal to 7.83 m/s.

Since t1 is zero, t2 is 9.35 s.

t2 = 9.35 s, v2 = 7.83 m/s

The elevator keeps the velocity constant for some time, and then, at some instant t3, it begins decelerating.

The velocity v3 that the elevator has at t3 is the same as the velocity at t2, i.e., 7.83 m/s.

t3 = unknown
v3 = 7.83 m/s

Finally, at an instant t4, 12.5 s after t3, the elevator comes to a stop.

The velocity v4 that the elevator has when it stops is equal to 0.

t4 β t3 = 12.5 s
v4 = 0

We need to find the average acceleration of the elevator for the acceleration phase, which starts at t1 and ends at t2, the deceleration phase, which starts at t3 and ends at t4, and from the moment the elevator starts accelerating to the moment it decelerates to a stop, i.e., from instant t1 to instant t4.

Let's begin by finding the average acceleration a12 between t1 and t2:

 a12 = v2 β v1 = 7.83 m/s β 0 = 7.83 m/s = 0.837 m/s2 t2 β t1 9.35 s β 0 9.35 s

Next, let's find the average acceleration a34 between t3 and t4:

 a34 = v4 β v3 = 0 β 7.83 m/s = β7.83 m/s = β0.626 m/s2 t4 β t3 12.5 s 12.5 s

In conclusion, let's find the average acceleration a14 between t1 and t4:

 a14 = v4 β v1 = 0 β 0 = 0 = 0 t4 β t1 t4 β t1 t4 β t1

Therefore, the elevator has a positive average acceleration (0.837 m/s2) for the acceleration phase, a negative average acceleration (β0.626 m/s2) for the deceleration phase, and a zero average acceleration from the moment it starts accelerating to the moment it decelerates to a stops.

## Velocity vs time graphs and average acceleration

A velocity vs time graph shows how the velocity of a particle changes over time.

Here's an example:

### How to find the average acceleration from a velocity vs time graph

A velocity vs time graph allows us to determine the velocity of a particle at any moment.

For example, in the velocity vs time graph shown above, at the instant t = 4 s, the particle has a velocity v = 60 m/s:

So, a velocity vs time graph gives us enough information to find the average acceleration of the particle for any interval of time.

For example, we can find the average acceleration between 4 s and 8 s from the above velocity vs time graph by following these 3 simple steps:

1. Look at the graph and determine the velocity v1 that the particle has at instant t1 = 4 s.
2. Look at the graph and determine the velocity v2 that the particle has at instant t2 = 8 s.
3. Calculate the average acceleration by dividing the change in velocity v2 β v1 by the interval of time t2 β t1.

t1 = 4 s, v1 = 60 m/s
t2 = 8 s, v2 = 80 m/s

Then, in step 3, we calculate the average acceleration:

 a12 = v2 β v1 = 80 m/s β 60 m/s = 20 m/s = 5 m/s2 t2 β t1 8 s β 4 s 4 s

### Average acceleration as the slope of a secant line to the velocity vs time graph

Let's consider a velocity vs time graph which shows how the velocity of a particle changed over time:

Now, let's take an instant t1 at which the particle has a velocity v1 and a subsequent instant t2 at which the particle has a velocity v2:

By looking at the graph, we can see that the change in velocity v2 β v1 = Ξv is the vertical distance between the points corresponding to t1 and t2 on the graph:

We can also see that the interval of time t2 β t1 = Ξt is the horizontal distance between the points corresponding to t1 and t2 on the graph:

Now, let's represent the secant line that passes through the points t1 and t2 on the graph:

The ratio of Ξv to Ξt is the ratio of the vertical change to the horizontal change between two points on the secant line. So, this ratio is equal to the slope m of the secant line:

 Ξv = m Ξt

We also know that this ratio is equal to the average acceleration between t1 and t2:

 Ξv = a Ξt

It follows that the slope m of the secant line is equal to the average acceleration a:

m = a

Therefore, the average acceleration between an instant t1 and an instant t2 is equal to the slope of the secant line that passes through the points t1 and t2 on the velocity vs time graph.

### How to determine the sign of the average acceleration from a velocity vs time graph

Determining the sign of the average acceleration from a velocity vs time graph turns out to be quite simple.

Let's go back to the same velocity vs time graph as above and consider an instant t1 at which the particle has a velocity v1 and a subsequent instant t2 at which the particle has a velocity v2:

Next, let's represent the secant line that passes through the points t1 and t2 on the velocity vs time graph and let's call ΞΈ the angle that it makes with the positive t-axis:

To determine whether the average acceleration between t1 and t2 is positive, negative, or zero, all we have to do is look at the sign of the angle ΞΈ:

1. ΞΈ > 0
Ξv > 0 β a > 0
2. ΞΈ < 0
Ξv < 0 β a < 0
3. ΞΈ = 0
Ξv = 0 β a = 0

So, the sign of the average acceleration between an instant t1 and an instant t2 is the same as that of the angle ΞΈ that the secant line, passing through the points t1 and t2 on the velocity vs time graph, makes with the positive t-axis.

## The change in velocity and the interval of time expressed in terms of the average acceleration

There many cases where we might want to express the change in velocity Ξv or the interval of time Ξt in terms of the average acceleration a.

We can easily do that by first remembering that the average acceleration a for an interval of time Ξt is equal to the ratio of the change in velocity Ξv, that occurs during Ξt, to Ξt itself:

 a = Ξv Ξt

Then, we can solve this equation for the change in velocity Ξv:

Ξv = aΞt

The way we should read this is that the change in velocity Ξv that occurs in an interval of time Ξt is equal to the average acceleration a for that interval of time multiplied by the interval of time itself.

We can also solve the above equation for the interval of time Ξt:

 Ξt = Ξv a

The way we should read this is that an interval of time Ξt is equal to the change in velocity Ξv that occurs during that interval of time divided by the average acceleration a for that interval of time.

## Example: Average acceleration of a rocket

After liftoff, a rocket arrives at a velocity of 1645 km/h with an average acceleration of 5.870 m/s2. Then, the rocket continues accelerating for 91.85 s, until the main engine is shut down, with an average acceleration of 20.32 m/s2. What is the average acceleration of the rocket from the moment it is about to lift off to the moment the main engine is shut down?

Let's consider the instant the rocket is about to lift off to be 0 and label it as t1. So, the velocity v1 at instant t1 is 0.

t1 = 0, v1 = 0

At a subsequent instant t2, the rocket arrives at a velocity of 1645 km/h. So, the velocity v2 at instant t2 is 1645 km/h.

The instant t2 is unknown. However, we know that the average acceleration between t1 and t2 is 5.870 m/s2.

t2 = unknown, v2 = 1645 km/h
a12 = 5.870 m/s2

Then, at instant t3, 91.85 s after t2, the main engine of the rocket is shut down.

The velocity v3 at the instant t3 is unknown. However, we know that the average acceleration between t2 and t3 is 20.32 m/s2.

t3 = unknown, v3 = unknown
t3 β t2 = 91.85 s
a23 = 20.32 m/s2

Our job now is to calculate the average acceleration of the rocket from the moment it is about to lift off to the moment the main engine is shut down, i.e., from instant t1 to instant t3.

The average acceleration a13 between t1 and t3 is

 a13 = v3 β v1 t3 β t1

Let's start by determining what v3 β v1 is.

We have already seen in the train example that the change in velocity v3 β v1 is equal to the sum of v2 β v1 and v3 β v2:

v3 β v1 = (v2 β v1) + (v3 β v2)

It is not difficult to find v2 β v1 since both v1 and v2 are known:

v2 β v1 = 1645 km/h β 0
v2 β v1 = 1645 km/h

But what about v3 β v2?

We know v2 but not v3.

Well, we know the interval of time between t2 and t3 as well as the average acceleration between t2 and t3:

t3 β t2 = 91.85 s
a23 = 20.32 m/s2

And since the average acceleration between t2 and t3 is expressed as

 a23 = v3 β v2 t3 β t2

Where both a23 and t3 β t2 are known, we can find the value of v3 β v2:

v3 β v2 = a23 (t3 β t2)
v3 β v2 = (20.32 m/s2) (91.85 s)
v3 β v2 = 1866 m/s

Now that v3 β v2 is known, we can calculate the value of v3 β v1:

v3 β v1 = (v2 β v1) + (v3 β v2)
v3 β v1 = 1645 km/h + 1866 m/s
 1645 km/h = 1645 m/s = 456.9 m/s 3.6
v3 β v1 = 456.9 m/s + 1866 m/s
v3 β v1 = 2323 m/s

So, we determined what v3 β v1 is.

Next, let's find t3 β t1.

t3 β t1 is equal to the sum of t2 β t1 and t3 β t2:

t3 β t1 = (t2 β t1) + (t3 β t2)

We know what t3 β t2 is:

t3 β t2 = 91.85 s

But we don't know what t2 β t1 is.

We only know t1 but not t2.

However, we know the change in velocity between t1 and t2 as well as the average acceleration between t1 and t2:

v2 β v1 = 1645 km/h = 456.9 m/s
a12 = 5.870 m/s2

And since the average acceleration between t1 and t2 is expressed as

 a12 = v2 β v1 t2 β t1

Where both a12 and v2 β v1 are known, we can find the value of t2 β t1:

(t2 β t1) a12 = v2 β v1
 t2 β t1 = v2 β v1 a12
 t2 β t1 = 456.9 m/s 5.870 m/s2
t2 β t1 = 77.84 s

Knowing what t2 β t1 is, we can calculate t3 β t1:

t3 β t1 = (t2 β t1) + (t3 β t2)
t3 β t1 = 77.84 s + 91.85 s
t3 β t1 = 169.69 s

Now that we know both v3 β v1 and t3 β t1, we can finally calculate the average acceleration a13:

 a13 = v3 β v1 t3 β t1
 a13 = 2323 m/s = 13.69 m/s2 169.69 s

Therefore, the average acceleration of the rocket, from the moment it is about to lift off to the moment the main engine is shut down, is 13.69 m/s2.

## Summary

• The average acceleration a between an instant t1 and an instant t2 is defined as the ratio of the change in velocity v2 β v1 to the interval of time t2 β t1:

 a = v2 β v1 = Ξv t2 β t1 Ξt
• The SI unit of average acceleration is meters per second per second ((m/s)/s) or more simply meters per second squared (m/s2).

• The sign of the average acceleration a is the same as the sign of the change in velocity Ξv.

1. Ξv > 0 β a > 0
2. Ξv < 0 β a < 0
3. Ξv = 0 β a = 0
• The average acceleration between an instant t1 and an instant t2 is equal to the slope of the secant line that passes through the points t1 and t2 on the velocity vs time graph.

The sign of the average acceleration is the same as the sign of the angle ΞΈ that the secant line makes with the positive t-axis.

1. ΞΈ > 0 β a > 0
2. ΞΈ < 0 β a < 0
3. ΞΈ = 0 β a = 0
• The change in velocity Ξv that occurs during an interval of time Ξt is equal to the average acceleration a for that interval of time multiplied by the interval of time itself:

Ξv = aΞt
• An interval of time Ξt is equal to the change in velocity Ξv that occurs during that interval of time divided by the average acceleration a for that interval of time:

 Ξt = Ξv a

## Exercises

### #1

When a rocket is launched, it reaches a velocity of 4140 km/h after only 95.0 s. Calculate the average acceleration that the rocket has for the first 95.0 s after launch.

### Solution

12.1 m/s2

#### How to arrive at the solution

t1 = 0, v1 = 0
t2 = 95.0 s, v2 = 4140 km/h
a12 = ?
 a12 = v2 β v1 = 4140 km/h β 0 = 4140 km/h t2 β t1 95.0 s β 0 95.0 s
 4140 km/h = 4140 m/s = 1150 m/s 3.6
 a12 = 1150 m/s = 12.1 m/s2 95.0 s

### #2

A car accelerates to a velocity of 81.0 km/h from rest in 5.60 s. It continues to accelerate for another 3.80 s, which causes its velocity to increase by 17.0 km/h. What are the average accelerations of the car from the moment it starts accelerating to the moment it reaches a velocity of 81.0 km/h, for the time interval during which it increments its velocity by 17.0 km/h, and from the start until it reaches its final velocity?

### Solution

4.02 m/s2; 1.24 m/s2; 2.89 m/s2

#### How to arrive at the solution

t1 = 0, v1 = 0
t2 = 5.60 s, v2 = 81.0 km/h
t3 β t2 = 3.80 s, v3 β v2 = 17.0 km/h
a12 = ?, a23 = ?, a13 = ?
 a12 = v2 β v1 = 81.0 km/h β 0 = 81.0 km/h t2 β t1 5.60 s β 0 5.60 s
 81.0 km/h = 81.0 m/s = 22.5 m/s 3.6
 a12 = 22.5 m/s = 4.02 m/s2 5.60 s
 a23 = v3 β v2 = 17.0 km/h t3 β t2 3.80 s
 17.0 km/h = 17.0 m/s = 4.72 m/s 3.6
 a23 = 4.72 m/s = 1.24 m/s2 3.80 s
 a13 = v3 β v1 t3 β t1
v3 β v1 = (v2 β v1) + (v3 β v2)
v3 β v1 = 81.0 km/h + 17.0 km/h
v3 β v1 = 98.0 km/h
t3 β t1 = (t2 β t1) + (t3 β t2)
t3 β t1 = 5.60 s + 3.80 s
t3 β t1 = 9.40 s
 a13 = v3 β v1 = 98.0 km/h t3 β t1 9.40 s
 98.0 km/h = 98.0 m/s = 27.2 m/s 3.6
 a13 = 27.2 m/s = 2.89 m/s2 9.40 s

### #3

A train departs from a station and speeds up to a velocity of 85.0 km/h in 30.0 s. It keeps moving with a constant velocity for a while. Then, when the train is about to arrive at the destination station, it brakes and successfully comes to a halt after 39.0 s. Find the average accelerations of the train for the speed up phase, for the braking phase, and from the moment it starts speeding up to moment it comes to a halt.

### Solution

0.787 m/s2; β0.605 m/s2; 0

#### How to arrive at the solution

t1 = 0, v1 = 0
t2 = 30.0 s, v2 = 85.0 km/h
t3 = unknown, v3 = 85.0 km/h
t4 β t3 = 39.0 s, v4 = 0
a12 = ?, a34 = ?, a14 = ?
 a12 = v2 β v1 = 85.0 km/h β 0 = 85.0 km/h t2 β t1 30.0 s β 0 30.0 s
 85.0 km/h = 85.0 m/s = 23.6 m/s 3.6
 a12 = 23.6 m/s = 0.787 m/s2 30.0 s
 a34 = v4 β v3 = 0 β 85.0 km/h = β85.0 km/h t4 β t3 39.0 s 39.0 s
 β85.0 km/h = β85.0 m/s = β23.6 m/s 3.6
 a34 = β23.6 m/s = β0.605 m/s2 39.0 s
 a14 = v4 β v1 = 0 β 0 = 0 = 0 t4 β t1 t4 β t1 t4 β t1

### #4

In a sprint, a runner has an average acceleration of 3.12 m/s2 for the first 2.40 s. After that, she keeps accelerating and further increases her velocity by 5.31 m/s with an average acceleration of 1.85 m/s2. Calculate the average acceleration of the runner for the entire acceleration period.

### Solution

2.43 m/s2

#### How to arrive at the solution

t1 = 0, v1 = 0
t2 = 2.40 s, v2 = unknown, a12 = 3.12 m/s2
t3 = unknown, v3 β v2 = 5.31 m/s, a23 = 1.85 m/s2
a13 = ?
 a13 = v3 β v1 t3 β t1
v3 β v1 = (v2 β v1) + (v3 β v2)
v3 β v2 = 5.31 m/s
t2 β t1 = 2.40 s β 0 = 2.40 s
a12 = 3.12 m/s2
 a12 = v2 β v1 t2 β t1
v2 β v1 = a12 (t2 β t1)
v2 β v1 = (3.12 m/s2) (2.40 s)
v2 β v1 = 7.49 m/s
v3 β v1 = (v2 β v1) + (v3 β v2)
v3 β v1 = 7.49 m/s + 5.31 m/s
v3 β v1 = 12.80 m/s
t3 β t1 = (t2 β t1) + (t3 β t2)
t2 β t1 = 2.40 s
v3 β v2 = 5.31 m/s
a23 = 1.85 m/s2
 a23 = v3 β v2 t3 β t2
(t3 β t2) a23 = v3 β v2
 t3 β t2 = v3 β v2 a23
 t3 β t2 = 5.31 m/s 1.85 m/s2
t3 β t2 = 2.87 s
t3 β t1 = (t2 β t1) + (t3 β t2)
t3 β t1 = 2.40 s + 2.87 s
t3 β t1 = 5.27 s
 a13 = v3 β v1 t3 β t1
 a13 = 12.80 m/s = 2.43 m/s2 5.27 s