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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**.

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

If at an instant t_{1} the particle has a velocity v_{1} and at a subsequent instant t_{2} it has a velocity v_{2}, we indicate the change in velocity between t_{1} and t_{2}, i.e., v_{2} β v_{1}, with Ξv (delta-v) :

v_{2} β v_{1} = Ξv

Also, we indicate the interval of time between t_{1} and t_{2}, i.e., t_{2} β t_{1}, with Ξt (delta-t) :

t_{2} β t_{1} = Ξt

The average acceleration a_{12} that the particle has between instant t_{1} and instant t_{2} is defined as the ratio of the change in velocity between t_{1} and t_{2}, i.e., v_{2} β v_{1}, to the interval of time between t_{1} and t_{2}, i.e., t_{2} β t_{1}.

Therefore, the **formula for average acceleration** is

a_{12} = | v_{2} β v_{1} | = | Ξv |

t_{2} β t_{1} | Ξt |

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

a = | Ξv |

Ξt |

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

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/s^{2}).

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:

v_{12} = | x_{2} β x_{1} | = | Ξx |

t_{2} β t_{1} | Ξt |

a_{12} = | v_{2} β v_{1} | = | Ξv |

t_{2} β t_{1} | Ξt |

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 t_{1}. So, the velocity v_{1} at instant t_{1} is 0.

t_{1} = 0, v_{1} = 0

We know that at a subsequent instant t_{2}, the car reaches a velocity of 100 km/h. So, the velocity v_{2} at instant t_{2} is 100 km/h.

Also, since the time that passes between t_{1} and t_{2} is 4.75 s, and t_{1} is zero, it follows that t_{2} is 4.75 s.

t_{2} = 4.75 s, v_{2} = 100 km/h

We want to find the average acceleration between t_{1} and t_{2}, which we label as a_{12}.

The average acceleration a_{12} between an instant t_{1} and an instant t_{2} is equal to the ratio of the change in velocity that occurs between t_{1} and t_{2} to the interval of time between t_{1} and t_{2}:

a_{12} = | v_{2} β v_{1} | = | 100 km/h β 0 | = | 100 km/h |

t_{2} β t_{1} | 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:

a_{12} = | 27.8 m/s | = 5.85 m/s^{2} |

4.75 s |

Thus, the average acceleration of the car is 5.85 m/s^{2}.

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 t_{1}. So, the velocity v_{1} at instant t_{1} is 0.

t_{1} = 0, v_{1} = 0

At a subsequent instant t_{2}, the train reaches a velocity of 40 km/h. So, the velocity v_{2} at instant t_{2} is 40 km/h.

The time that passes between the instant t_{1} and the instant t_{2} is 13 s, and since t_{1} is zero, it follows that t_{2} is 13 s.

t_{2} = 13 s, v_{2} = 40 km/h

At an instant t_{3}, 25 s after t_{2}, the train has incremented its velocity by 35 km/h. So, at instant t_{3}, the train has a velocity v_{3} which is greater than v_{2} by 35 km/h.

v_{3} β v_{2} = 35 km/h

t_{3} β t_{2} = 25 s

Our task is to calculate the average acceleration of the train for the first 13 s, which start at t_{1} and end at t_{2}, the following 25 s, which start at t_{2} and end at t_{3}, and the overall average acceleration, i.e., the average acceleration from the starting instant t_{1} to the final instant t_{3}.

Let's get started by calculating the average acceleration a_{12} between t_{1} and t_{2}:

a_{12} = | v_{2} β v_{1} | = | 40 km/h β 0 | = | 40 km/h |

t_{2} β t_{1} | 13 s β 0 | 13 s |

40 km/h = | 40 | m/s = 11 m/s |

3.6 |

a_{12} = | 11 m/s | = 0.85 m/s^{2} |

13 s |

Next, let's calculate the average acceleration a_{23} between t_{2} and t_{3}:

a_{23} = | v_{3} β v_{2} | = | 35 km/h |

t_{3} β t_{2} | 25 s |

35 km/h = | 35 | m/s = 9.7 m/s |

3.6 |

a_{23} = | 9.7 m/s | = 0.39 m/s^{2} |

25 s |

Last but not least, let's find the average acceleration a_{13} between t_{1} and t_{3}:

a_{13} = | v_{3} β v_{1} |

t_{3} β t_{1} |

How can we determine what v_{3} β v_{1} 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 v_{3} β v_{1} is equal to the sum of v_{2} β v_{1} and v_{3} β v_{2}:

v_{3} β v_{1} = (v_{2} β v_{1}) + (v_{3} β v_{2})

v_{3} β v_{1} = 40 km/h + 35 km/h

v_{3} β v_{1} = 75 km/h

With that in mind, it shouldn't be difficult to see that the interval of time t_{3} β t_{1} is equal to t_{2} β t_{1} and t_{3} β t_{2}:

t_{3} β t_{1} = (t_{2} β t_{1}) + (t_{3} β t_{2})

t_{3} β t_{1} = 13 s + 25 s

t_{3} β t_{1} = 38 s

Thus, the average acceleration a_{13} is

a_{13} = | 75 km/h |

38 s |

75 km/h = | 75 | m/s = 21 m/s |

3.6 |

a_{13} = | 21 m/s | = 0.55 m/s^{2} |

38 s |

Note that although in some cases the overall average acceleration a_{13} might be equal to the arithmetic mean of the average acceleration a_{12} and the average acceleration a_{23}, 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 a_{12} and a_{23}:

a_{12} + a_{23} | = | 0.85 m/s^{2} + 0.39 m/s^{2} | = 0.62 m/s^{2} |

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.

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

a = | Ξv |

Ξt |

and the interval of time Ξt is always positive, it follows that the sign of the average acceleration a is the same as the sign of the change in velocity Ξv.

The change in velocity Ξv 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:

- Ξv > 0 β a > 0
- Ξv < 0 β a < 0
- Ξv = 0 β a = 0

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 t_{1}. Thus, the velocity v_{1} at instant t_{1} is 0.

t_{1} = 0, v_{1} = 0

At a subsequent instant t_{2}, 9.35 s after t_{1}, the elevator reaches a velocity of 7.83 m/s. Therefore, the velocity v_{2} at instant t_{2} is equal to 7.83 m/s.

Since t_{1} is zero, t_{2} is 9.35 s.

t_{2} = 9.35 s, v_{2} = 7.83 m/s

The elevator keeps the velocity constant for some time, and then, at some instant t_{3}, it begins decelerating.

The velocity v_{3} that the elevator has at t_{3} is the same as the velocity at t_{2}, i.e., 7.83 m/s.

t_{3} = unknown

v_{3} = 7.83 m/s

Finally, at an instant t_{4}, 12.5 s after t_{3}, the elevator comes to a stop.

The velocity v_{4} that the elevator has when it stops is equal to 0.

t_{4} β t_{3} = 12.5 s

v_{4} = 0

We need to find the average acceleration of the elevator for the acceleration phase, which starts at t_{1} and ends at t_{2}, the deceleration phase, which starts at t_{3} and ends at t_{4}, and from the moment the elevator starts accelerating to the moment it decelerates to a stop, i.e., from instant t_{1} to instant t_{4}.

Let's begin by finding the average acceleration a_{12} between t_{1} and t_{2}:

a_{12} = | v_{2} β v_{1} | = | 7.83 m/s β 0 | = | 7.83 m/s | = 0.837 m/s^{2} |

t_{2} β t_{1} | 9.35 s β 0 | 9.35 s |

Next, let's find the average acceleration a_{34} between t_{3} and t_{4}:

a_{34} = | v_{4} β v_{3} | = | 0 β 7.83 m/s | = | β7.83 m/s | = β0.626 m/s^{2} |

t_{4} β t_{3} | 12.5 s | 12.5 s |

In conclusion, let's find the average acceleration a_{14} between t_{1} and t_{4}:

a_{14} = | v_{4} β v_{1} | = | 0 β 0 | = | 0 | = 0 |

t_{4} β t_{1} | t_{4} β t_{1} | t_{4} β t_{1} |

Therefore, the elevator has a positive average acceleration (0.837 m/s^{2}) for the acceleration phase, a negative average acceleration (β0.626 m/s^{2}) for the deceleration phase, and a zero average acceleration from the moment it starts accelerating to the moment it decelerates to a stops.

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

Here's an example:

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:

- Look at the graph and determine the velocity v
_{1}that the particle has at instant t_{1}= 4 s. - Look at the graph and determine the velocity v
_{2}that the particle has at instant t_{2}= 8 s. - Calculate the average acceleration by dividing the change in velocity v
_{2}β v_{1}by the interval of time t_{2}β t_{1}.

Let's start with step 1 and step 2:

t_{1} = 4 s, v_{1} = 60 m/s

t_{2} = 8 s, v_{2} = 80 m/s

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

a_{12} = | v_{2} β v_{1} | = | 80 m/s β 60 m/s | = | 20 m/s | = 5 m/s^{2} |

t_{2} β t_{1} | 8 s β 4 s | 4 s |

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 t_{1} at which the particle has a velocity v_{1} and a subsequent instant t_{2} at which the particle has a velocity v_{2}:

By looking at the graph, we can see that the change in velocity v_{2} β v_{1} = Ξv is the vertical distance between the points corresponding to t_{1} and t_{2} on the graph:

We can also see that the interval of time t_{2} β t_{1} = Ξt is the horizontal distance between the points corresponding to t_{1} and t_{2} on the graph:

Now, let's represent the secant line that passes through the points t_{1} and t_{2} on the graph:

The ratio of Ξv to Ξt 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:

Ξv | = m |

Ξt |

We also know that this ratio is equal to the average acceleration between t_{1} and t_{2}:

Ξv | = a |

Ξt |

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 t_{1} and an instant t_{2} is equal to the slope of the secant line that passes through the points t_{1} and t_{2} on the 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 t_{1} at which the particle has a velocity v_{1} and a subsequent instant t_{2} at which the particle has a velocity v_{2}:

Next, let's represent the secant line that passes through the points t_{1} and t_{2} 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 t_{1} and t_{2} is positive, negative, or zero, all we have to do is look at the sign of the angle ΞΈ:

- ΞΈ > 0Ξv > 0 β a > 0
- ΞΈ < 0Ξv < 0 β a < 0
- ΞΈ = 0Ξv = 0 β a = 0

So, the sign of the average acceleration between an instant t_{1} and an instant t_{2} is the same as that of the angle ΞΈ that the secant line, passing through the points t_{1} and t_{2} on the velocity vs time graph, makes with the positive t-axis.

There many cases where we might want to express the change in velocity Ξv or the interval of time Ξt 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 Ξt is equal to the ratio of the change in velocity Ξv, that occurs during Ξt, to Ξt itself:

a = | Ξv |

Ξt |

Then, we can solve this equation for the change in velocity Ξv:

Ξv = aΞt

The way we should read this is that the change in velocity Ξv that occurs in an interval of time Ξt 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 Ξt:

Ξt = | Ξv |

a |

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

After liftoff, a rocket arrives at a velocity of 1645 km/h with an average acceleration of 5.870 m/s^{2}. Then, the rocket continues accelerating for 91.85 s, until the main engine is shut down, with an average acceleration of 20.32 m/s^{2}. 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 t_{1}. So, the velocity v_{1} at instant t_{1} is 0.

t_{1} = 0, v_{1} = 0

At a subsequent instant t_{2}, the rocket arrives at a velocity of 1645 km/h. So, the velocity v_{2} at instant t_{2} is 1645 km/h.

The instant t_{2} is unknown. However, we know that the average acceleration between t_{1} and t_{2} is 5.870 m/s^{2}.

t_{2} = unknown, v_{2} = 1645 km/h

a_{12} = 5.870 m/s^{2}

Then, at instant t_{3}, 91.85 s after t_{2}, the main engine of the rocket is shut down.

The velocity v_{3} at the instant t_{3} is unknown. However, we know that the average acceleration between t_{2} and t_{3} is 20.32 m/s^{2}.

t_{3} = unknown, v_{3} = unknown

t_{3} β t_{2} = 91.85 s

a_{23} = 20.32 m/s^{2}

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 t_{1} to instant t_{3}.

The average acceleration a_{13} between t_{1} and t_{3} is

a_{13} = | v_{3} β v_{1} |

t_{3} β t_{1} |

Let's start by determining what v_{3} β v_{1} is.

We have already seen in the train example that the change in velocity v_{3} β v_{1} is equal to the sum of v_{2} β v_{1} and v_{3} β v_{2}:

v_{3} β v_{1} = (v_{2} β v_{1}) + (v_{3} β v_{2})

It is not difficult to find v_{2} β v_{1} since both v_{1} and v_{2} are known:

v_{2} β v_{1} = 1645 km/h β 0

v_{2} β v_{1} = 1645 km/h

But what about v_{3} β v_{2}?

We know v_{2} but not v_{3}.

Well, we know the interval of time between t_{2} and t_{3} as well as the average acceleration between t_{2} and t_{3}:

t_{3} β t_{2} = 91.85 s

a_{23} = 20.32 m/s^{2}

And since the average acceleration between t_{2} and t_{3} is expressed as

a_{23} = | v_{3} β v_{2} |

t_{3} β t_{2} |

Where both a_{23} and t_{3} β t_{2} are known, we can find the value of v_{3} β v_{2}:

v_{3} β v_{2} = a_{23} (t_{3} β t_{2})

v_{3} β v_{2} = (20.32 m/s^{2}) (91.85 s)

v_{3} β v_{2} = 1866 m/s

Now that v_{3} β v_{2} is known, we can calculate the value of v_{3} β v_{1}:

v_{3} β v_{1} = (v_{2} β v_{1}) + (v_{3} β v_{2})

v_{3} β v_{1} = 1645 km/h + 1866 m/s

1645 km/h = | 1645 | m/s = 456.9 m/s |

3.6 |

v_{3} β v_{1} = 456.9 m/s + 1866 m/s

v_{3} β v_{1} = 2323 m/s

So, we determined what v_{3} β v_{1} is.

Next, let's find t_{3} β t_{1}.

t_{3} β t_{1} is equal to the sum of t_{2} β t_{1} and t_{3} β t_{2}:

t_{3} β t_{1} = (t_{2} β t_{1}) + (t_{3} β t_{2})

We know what t_{3} β t_{2} is:

t_{3} β t_{2} = 91.85 s

But we don't know what t_{2} β t_{1} is.

We only know t_{1} but not t_{2}.

However, we know the change in velocity between t_{1} and t_{2} as well as the average acceleration between t_{1} and t_{2}:

v_{2} β v_{1} = 1645 km/h = 456.9 m/s

a_{12} = 5.870 m/s^{2}

And since the average acceleration between t_{1} and t_{2} is expressed as

a_{12} = | v_{2} β v_{1} |

t_{2} β t_{1} |

Where both a_{12} and v_{2} β v_{1} are known, we can find the value of t_{2} β t_{1}:

(t_{2} β t_{1}) a_{12} = v_{2} β v_{1}

t_{2} β t_{1} = | v_{2} β v_{1} |

a_{12} |

t_{2} β t_{1} = | 456.9 m/s |

5.870 m/s^{2} |

t_{2} β t_{1} = 77.84 s

Knowing what t_{2} β t_{1} is, we can calculate t_{3} β t_{1}:

t_{3} β t_{1} = (t_{2} β t_{1}) + (t_{3} β t_{2})

t_{3} β t_{1} = 77.84 s + 91.85 s

t_{3} β t_{1} = 169.69 s

Now that we know both v_{3} β v_{1} and t_{3} β t_{1}, we can finally calculate the average acceleration a_{13}:

a_{13} = | v_{3} β v_{1} |

t_{3} β t_{1} |

a_{13} = | 2323 m/s | = 13.69 m/s^{2} |

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/s^{2}.

The average acceleration a between an instant t

_{1}and an instant t_{2}is defined as the ratio of the change in velocity v_{2}β v_{1}to the interval of time t_{2}β t_{1}:a = v _{2}β v_{1}= Ξv t _{2}β t_{1}Ξt The SI unit of average acceleration is meters per second per second ((m/s)/s) or more simply meters per second squared (m/s

^{2}).The sign of the average acceleration a is the same as the sign of the change in velocity Ξv.

- Ξv > 0 β a > 0
- Ξv < 0 β a < 0
- Ξv = 0 β a = 0

The average acceleration between an instant t

_{1}and an instant t_{2}is equal to the slope of the secant line that passes through the points t_{1}and t_{2}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.

- ΞΈ > 0 β a > 0
- ΞΈ < 0 β a < 0
- ΞΈ = 0 β a = 0

The change in velocity Ξv that occurs during an interval of time Ξt is equal to the average acceleration a for that interval of time multiplied by the interval of time itself:

Ξv = aΞtAn interval of time Ξt is equal to the change in velocity Ξv that occurs during that interval of time divided by the average acceleration a for that interval of time:

Ξt = Ξv a

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.

12.1 m/s^{2}

t_{1} = 0, v_{1} = 0

t_{2} = 95.0 s, v_{2} = 4140 km/h

a_{12} = ?

a_{12} = | v_{2} β v_{1} | = | 4140 km/h β 0 | = | 4140 km/h |

t_{2} β t_{1} | 95.0 s β 0 | 95.0 s |

4140 km/h = | 4140 | m/s = 1150 m/s |

3.6 |

a_{12} = | 1150 m/s | = 12.1 m/s^{2} |

95.0 s |

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?

4.02 m/s^{2}; 1.24 m/s^{2}; 2.89 m/s^{2}

t_{1} = 0, v_{1} = 0

t_{2} = 5.60 s, v_{2} = 81.0 km/h

t_{3} β t_{2} = 3.80 s, v_{3} β v_{2} = 17.0 km/h

a_{12} = ?, a_{23} = ?, a_{13} = ?

a_{12} = | v_{2} β v_{1} | = | 81.0 km/h β 0 | = | 81.0 km/h |

t_{2} β t_{1} | 5.60 s β 0 | 5.60 s |

81.0 km/h = | 81.0 | m/s = 22.5 m/s |

3.6 |

a_{12} = | 22.5 m/s | = 4.02 m/s^{2} |

5.60 s |

a_{23} = | v_{3} β v_{2} | = | 17.0 km/h |

t_{3} β t_{2} | 3.80 s |

17.0 km/h = | 17.0 | m/s = 4.72 m/s |

3.6 |

a_{23} = | 4.72 m/s | = 1.24 m/s^{2} |

3.80 s |

a_{13} = | v_{3} β v_{1} |

t_{3} β t_{1} |

v_{3} β v_{1} = (v_{2} β v_{1}) + (v_{3} β v_{2})

v_{3} β v_{1} = 81.0 km/h + 17.0 km/h

v_{3} β v_{1} = 98.0 km/h

t_{3} β t_{1} = (t_{2} β t_{1}) + (t_{3} β t_{2})

t_{3} β t_{1} = 5.60 s + 3.80 s

t_{3} β t_{1} = 9.40 s

a_{13} = | v_{3} β v_{1} | = | 98.0 km/h |

t_{3} β t_{1} | 9.40 s |

98.0 km/h = | 98.0 | m/s = 27.2 m/s |

3.6 |

a_{13} = | 27.2 m/s | = 2.89 m/s^{2} |

9.40 s |

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.

0.787 m/s^{2}; β0.605 m/s^{2}; 0

t_{1} = 0, v_{1} = 0

t_{2} = 30.0 s, v_{2} = 85.0 km/h

t_{3} = unknown, v_{3} = 85.0 km/h

t_{4} β t_{3} = 39.0 s, v_{4} = 0

a_{12} = ?, a_{34} = ?, a_{14} = ?

a_{12} = | v_{2} β v_{1} | = | 85.0 km/h β 0 | = | 85.0 km/h |

t_{2} β t_{1} | 30.0 s β 0 | 30.0 s |

85.0 km/h = | 85.0 | m/s = 23.6 m/s |

3.6 |

a_{12} = | 23.6 m/s | = 0.787 m/s^{2} |

30.0 s |

a_{34} = | v_{4} β v_{3} | = | 0 β 85.0 km/h | = | β85.0 km/h |

t_{4} β t_{3} | 39.0 s | 39.0 s |

β85.0 km/h = | β85.0 | m/s = β23.6 m/s |

3.6 |

a_{34} = | β23.6 m/s | = β0.605 m/s^{2} |

39.0 s |

a_{14} = | v_{4} β v_{1} | = | 0 β 0 | = | 0 | = 0 |

t_{4} β t_{1} | t_{4} β t_{1} | t_{4} β t_{1} |

In a sprint, a runner has an average acceleration of 3.12 m/s^{2} 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/s^{2}. Calculate the average acceleration of the runner for the entire acceleration period.

2.43 m/s^{2}

t_{1} = 0, v_{1} = 0

t_{2} = 2.40 s, v_{2} = unknown, a_{12} = 3.12 m/s^{2}

t_{3} = unknown, v_{3} β v_{2} = 5.31 m/s, a_{23} = 1.85 m/s^{2}

a_{13} = ?

a_{13} = | v_{3} β v_{1} |

t_{3} β t_{1} |

v_{3} β v_{1} = (v_{2} β v_{1}) + (v_{3} β v_{2})

v_{3} β v_{2} = 5.31 m/s

t_{2} β t_{1} = 2.40 s β 0 = 2.40 s

a_{12} = 3.12 m/s^{2}

a_{12} = | v_{2} β v_{1} |

t_{2} β t_{1} |

v_{2} β v_{1} = a_{12} (t_{2} β t_{1})

v_{2} β v_{1} = (3.12 m/s^{2}) (2.40 s)

v_{2} β v_{1} = 7.49 m/s

v_{3} β v_{1} = (v_{2} β v_{1}) + (v_{3} β v_{2})

v_{3} β v_{1} = 7.49 m/s + 5.31 m/s

v_{3} β v_{1} = 12.80 m/s

t_{3} β t_{1} = (t_{2} β t_{1}) + (t_{3} β t_{2})

t_{2} β t_{1} = 2.40 s

v_{3} β v_{2} = 5.31 m/s

a_{23} = 1.85 m/s^{2}

a_{23} = | v_{3} β v_{2} |

t_{3} β t_{2} |

(t_{3} β t_{2}) a_{23} = v_{3} β v_{2}

t_{3} β t_{2} = | v_{3} β v_{2} |

a_{23} |

t_{3} β t_{2} = | 5.31 m/s |

1.85 m/s^{2} |

t_{3} β t_{2} = 2.87 s

t_{3} β t_{1} = (t_{2} β t_{1}) + (t_{3} β t_{2})

t_{3} β t_{1} = 2.40 s + 2.87 s

t_{3} β t_{1} = 5.27 s

a_{13} = | v_{3} β v_{1} |

t_{3} β t_{1} |

a_{13} = | 12.80 m/s | = 2.43 m/s^{2} |

5.27 s |