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In this Physics tutorial, you will learn:

- What are motion maps and what are they used for?
- What is acceleration and how can we calculate it?
- What is deceleration and how it relates to the acceleration?
- What is constant and non-constant acceleration?
- What is the average and instantaneous acceleration and when should we use them?
- What is gravitational acceleration and what is its value near the Earth surface?

**What do you do if you realize you are late for a meeting? Do you still walk or drive at the same speed as before?**

**Are you able to keep running at the same speed during a long race? What happens to your speed towards the end of the race? Why?**

In this tutorial we will discuss about these scenarios and the quantities (both old and new) involved in them.

Before starting with the topic, it is worth explaining the meaning of "motion maps."

In Kinematics, Motion Maps are a useful visual tool for analysing and demonstrating what we know about how an object moves. In general, a motion map can represent the position, velocity, and acceleration of an object at various clock readings (we haven't discussed the acceleration yet, but we will cover it in this tutorial, so it is better including the concept of acceleration in the motion map explanation).

A motion map has many similarities with a motion graph. They both have a origin (reference point), a positive and a negative direction and they both include units. The only difference is that a motion graph doesn't show any object but only numbers. On the other hand, a motion map shows the moving object and the way how it is moving is shown through arrows. Longer the arrows, faster the object is moving.

The figure below represents a simple motion map.

The black dots represent an object "photographed" in equal intervals of time (clock readings). It is easy to notice that the object (the black dot) is making a uniform motion (it has a constant velocity) because the distance between two consecutive dots (which in fact represent the position of the object at every two consecutive equal time intervals) is always the same (5 units). Therefore, we can appoint a velocity vector (arrow) to each black dot whose length (in this case) is always the same. The arrow is also inserted to show the moving direction as it's not for sure the object is always moving in the positive direction. Look at the figure below.

We can understand this is a uniform motion judging on the arrows' length only. In this way, we don't have to check whether the object has moved the same number of units at each time interval.

We can appoint numbers to the units shown in the motion map. We can illustrate the time intervals with numerical values as well. For example, if one square represents 1 m and the time interval between two consecutive dots (the clock) is 0.5 s, we can find the following values:

Displacement at every time interval = ∆x*⃗* = 5 m

Total Displacement = ∆x*⃗*_{tot} = 8 × 5 m = 40 m

Time interval (clock) = ∆t = 0.5 s

Total time elapsed = t_{tot} = 8 × 0.5 s = 4 s

Average velocity = < v*⃗* > = *∆x**⃗*_{tot}*/**t*_{tot}= *40 m**/**4s*= 10 *m**/**s*

Instantaneous velocity = v*⃗*= *∆x**⃗**/**∆t *= *5m**/**0.5s*= 10 *m**/**s*

Total Displacement = ∆x

Time interval (clock) = ∆t = 0.5 s

Total time elapsed = t

Average velocity = < v

Instantaneous velocity = v

(We have considered a single distance unit and consequently a single time interval when calculating the instantaneous velocity).

Since the above motion map represents a uniform motion (motion done at the same velocity), we can draw an important conclusion based on the above findings:

*"In a uniform motion, the values of average and instantaneous velocity are equal."*

This conclusion is drawn with the help of the motion map discussed above.

Now, let's use the help of motion maps to analyse situations involving objects that are not moving at the same velocity (and speed). Look at the figure below:

The ball is photographed in equal time intervals during its motion and its positions at each instant are shown above. It is obvious the ball is not moving at the same (constant) velocity as the distance between two consecutive positions of the ball increases while shifting due right. Therefore, we say "the ball is speeding up." In scientific terms, we say "the ball is accelerating." Thus, it is obvious "speeding up" and "accelerating" are similar terms that express the same physical process, i.e. "an increase in distance travelled at equal time intervals" or more precisely, "an increase in velocity at equal time intervals."

From the above discussion, we can formulate the definition of Acceleration as one of the basic Kinematic quantities:

**"Acceleration is the rate of velocity change per unit of time."**

The term "Rate" means "something changing with time." Therefore, if the change in velocity ∆v*⃗* = v*⃗*_{2} - v*⃗*_{1} occurs during a time interval ∆t, we obtain the formula of acceleration a*⃗*:

a*⃗* = *∆v**⃗**/**∆t*

=*v**⃗*_{2} - v*⃗*_{1}*/**∆t*

=

Since velocity is measured in ** m/s** and time in s, the unit of acceleration is

Unit of acceleration = [ *m**/**s**/**s* ] = [*m**/**s*^{2}]

As discussed in the previous tutorials "Speed and Velocity in One Dimension" and "Speed and Velocity in Two and Three Dimensions", velocity is a vector quantity because it involves direction. Therefore, acceleration is a vector quantity as well, as it is obtained by dividing a vector (the change in velocity) by a scalar (the time interval).

If we take the time intervals equal to 1s each in the motion map shown above, we can use numbers to understand better the meaning of acceleration. Thus, during the first second, the ball has moved due right by 1m, during the next seconds by 2m, during the third second by 3m and so on. This means the (average) velocity during the first second is 1m/1s = 1** m/s**, during the next second it is 2m/1s = 2

We have:

v*⃗*_{1} - v*⃗*_{0} = 1 *m**/**s* - 0 *m**/**s* = 1 *m**/**s*

v*⃗*_{2} - v*⃗*_{1} = 2 *m**/**s* - 1 *m**/**s* = 1 *m**/**s*

v*⃗*_{3} - v*⃗*_{2} = 3 *m**/**s* - 2 *m**/**s* = 1 *m**/**s*

v*⃗*_{4} - v*⃗*_{3} = 4 *m**/**s* - 3 *m**/**s* = 1 *m**/**s*

v

v

v

and so on. Thus, since each time interval is ∆t=1s, we have for the acceleration in each interval:

a*⃗*_{1} = *v**⃗*_{1} - v*⃗*_{0}*/**∆t* = *1 **m**/**s* - 0 *m**/**s**/**1s*= *1 **m**/**s**/**1s*= 1 *m**/**s*^{2}

a*⃗*_{2} = *v**⃗*_{2} - v*⃗*_{1}*/**∆t* = *2 **m**/**s* - 1 *m**/**s**/**1s*= *1 **m**/**s**/**1s*= 1 *m**/**s*^{2}

a*⃗*_{3} = *v**⃗*_{3} - v*⃗*_{2}*/**∆t* = *3 **m**/**s* - 2 *m**/**s**/**1s*= *1 **m**/**s**/**1s*= 1 *m**/**s*^{2}

a*⃗*_{4} = *v**⃗*_{4} - v*⃗*_{3}*/**∆t* = *4 **m**/**s* - 3 *m**/**s**/**1s*= *1 **m**/**s**/**1s*= 1 *m**/**s*^{2}

a

a

a

and so on. Therefore, it is obvious that in this example the increase in velocity occurred at the same rate. This means the acceleration - or as otherwise known, the rate of velocity change (here, the rate of velocity increase) - is constant (here, a = 1m/s^{2} in each interval). Hence, the situation described in the motion map discussed above, represents a "motion with constant acceleration."

Look at the motion map again. Now it includes the numerical values to make clearer the concept of acceleration and the reason why it is constant in the specific case.

On the other hand, if the object slows down, we say it is decelerating. Deceleration is the opposite of acceleration, i.e. it is a "negative acceleration." In such cases, the velocity decreases with time, i.e. the values of velocity in the successive intervals are smaller than those in the previous ones. This means the change in velocity is negative as in the expression ∆v*⃗* = v*⃗*_{i + 1} - v*⃗*_{i} (where i represents the number of the interval considered), the successive value of velocity (v*⃗*_{i + 1}) is smaller than the previous one (v*⃗*_{i}). Thus, when dividing a negative number (∆v*⃗*) by a positive one (Δt) the result is negative.

Let's take a look at another motion map to make this concept clear.

The arrow lengths become shorter and shorter until there is no arrow appearing in the figure. This means the ball stops at that position. If we appoint some values to the time intervals (for convenience, we can take them as Δt = 1s), and one unit in the graph is equal to 1m, we have for the average velocity during the first second:

v*⃗*_{1} = *8m**/**1s* = 8 *m**/**s*

In the next second, we have

v*⃗*_{2} = *6m**/**1s* = 6 *m**/**s*

In the third second, we have

v*⃗*_{3} = *4m**/**1s* = 4 *m**/**s*

In the fourth second, we have

v*⃗*_{4} = *2m**/**1s* = 2 *m**/**s*

and in the fifth (and the last) second, we have

v*⃗*_{5} = *0m**/**1s* = 0 *m**/**s*

Thus, the acceleration in each interval is

a*⃗* = *v**⃗*_{2}-v*⃗*_{1}*/**∆t* = *6 **m**/**s* - 8 *m**/**s**/**1s*= *-2 **m**/**s**/**1s*= -2 *m**/**s*^{2}

a*⃗* = *v**⃗*_{2}-v*⃗*_{1}*/**∆t* = *4 **m**/**s* - 6 *m**/**s**/**1s*= *-2 **m**/**s**/**1s*= -2 *m**/**s*^{2}

a*⃗* = *v**⃗*_{2}-v*⃗*_{1}*/**∆t* = *2 **m**/**s* - 4 *m**/**s**/**1s*= *-2 **m**/**s**/**1s*= -2 *m**/**s*^{2}

a*⃗* = *v**⃗*_{2}-v*⃗*_{1}*/**∆t* = *0 **m**/**s* - 2 *m**/**s**/**1s*= *-2 **m**/**s**/**1s*= -2 *m**/**s*^{2}

a

a

a

From the results obtained, it is clear this is a uniformly decelerated motion as the acceleration is constant and negative.

As we have stated in the tutorial Speed and Velocity in One Direction, rarely occurs that an object keeps moving at the same velocity during the entire motion. The same thing can be said for the constant acceleration as well. It is quite impossible an object keeps accelerating at the same rate during its motion, as many other factors such as the resistive forces caused by the air etc., are not constant. We will see in the Physics tutorial Types of Forces II. Resistive Forces (Frictional Force. Drag). Terminal Velocity that air resistance increases when an object is moving faster than before. This brings a decrease in the acceleration until it becomes zero. As a result, the object successively will move at constant velocity.

Another similar example consists in a scenario in which a driver presses the brakes near the traffic lights and after a while (when the green light turns on), he presses the gas pedal. It is obvious this is an irregular motion although it may be rectilinear. This irregularity results in the impossibility to consider the motion as a whole. In this case, we can use one of the two following approaches:

- Splitting the motion in small regular sections and finding the required quantities (here the acceleration) one by one before making any generalization, or
- Considering the motion as a whole and calculating only the average values of the entire motion. In this tutorial, we are interested only in the average acceleration which is obtained by the following equation:

< a*⃗* > = *∆v**⃗*_{tot}*/**t*_{tot}

The second method is much shorter and easier because we need to know only three quantities: the two values of velocity (the initial and the final velocity) and the total time taken.

On the other hand, the instantaneous acceleration shows the acceleration at a given instant, similarly to the instantaneous velocity we discussed in the previous tutorials. Thus, if we are required to find the instantaneous acceleration in a certain instant t, we need to consider a very narrow time interval Δt around the given instant and calculating the respective velocities v*⃗*_{1} and v*⃗*_{2} for the two bordering points of the interval. Then, using the equation

a*⃗* = *∆v**⃗**/**∆t* = *v**⃗*_{2} - v*⃗*_{1}*/**t*_{1} - t_{1}

we find the value of the instantaneous acceleration.

To help understand this point, a velocity vs time graph is shown below.

Let's suppose we are required to find the average acceleration t = 1.0 s. For this, we have to consider two surrounding points close to the given coordinate. For example, we can take t_{1} = 0.6 s and t_{2} = 1.4 s. Therefore, Δt = t_{2} - _{1} = 1.4 s - 0.6 s = 0.8 s.

From the graph, it is easy to find that the respective values of velocity for these time values are v*⃗*_{1} = 11 ** m/s** and v

a*⃗* = *∆v**⃗**/**∆t*

=*v**⃗*_{2}-v*⃗*_{1}*/**t*_{1} - t_{1}

=*13 **m**/**s* - 11 *m**/**s**/**1.4 s - 0.6 s*

=*2 **m**/**s**/**0.8 s*

= 2.5*m**/**s*^{2}

=

=

=

= 2.5

An object starts moving from rest and it reaches a velocity of 16 ** m/s** in 8 seconds. Then, it starts decelerating for the next 10 seconds until it comes to rest again.

**Calculate:**

- The average acceleration during the first interval (during the first 8s)
- The average acceleration during the second interval (during the next 10s)
- The total average acceleration

The initial velocity is usually written in problems as u*⃗* or v*⃗*_{0}; it is rarely written as v*⃗*_{1} or v*⃗*_{i}. Also, the final velocity is often written as v*⃗* instead of v*⃗*_{2} or v*⃗*_{f}. Therefore, we can write the following values as clues:

u = 0

v1 = 16*m**/**s*

Δt1 = 8s

v = 0

Δt2 = 10s

v1 = 16

Δt1 = 8s

v = 0

Δt2 = 10s

We have to find < a*⃗*_{1} >, < a*⃗*_{2} > and < a*⃗*_{tot} > because we are not sure about how the object has moved; we only know the three values of velocity. No information regarding any possible motion with constant acceleration is provides. For this reason, we can only calculate the average values of acceleration, not the instantaneous acceleration in any section of motion.

The average acceleration during the first interval is:

< a*⃗*_{1} > = *v**⃗*_{1}-u*⃗**/**∆t*_{1}

=*16 **m**/**s* - 0 *m**/**s**/**8s*

= 2*m**/**s*^{2}

=

= 2

The average acceleration during the second interval is:

< a*⃗*_{2} > = *v**⃗*-v*⃗*_{1}*/**∆t*_{2}

=*0 **m**/**s* - 16 *m**/**s**/**10s*

= -1.6*m**/**s*^{2}

=

= -1.6

Since the initial and final velocities are both zero, the total average acceleration is:

< a*⃗*_{tot} > = *∆v**⃗*_{tot}*/**t*_{tot}

=*v**⃗* - v*⃗*_{0}*/**t*_{tot}

=*0 - 0**/**8s + 10s*

=*0**/**18s* = 0

=

=

=

This is obvious because the positive acceleration in the first interval is cancelled out by the negative acceleration in the second interval.

Be careful to avoid taking the arithmetic mean of the two accelerations as in this case you get a wrong result. The only case in which it is allowed to calculate the average acceleration by finding the arithmetic mean is when the time intervals are equal.

It is a known fact that Earth causes an attraction effect on objects near its surface. This attraction effect causes a downward acceleration as object move faster and faster as they approach the Earth surface. If we neglect the opposing effect of air (air resistance), we consider this acceleration as constant. It is known as the "gravitational acceleration" or the "acceleration of free fall." It is denoted by g*⃗* and has a value 9.81 m/s^{2} near the Earth surface. When moving away from the Earth surface, the value of gravitational acceleration decreases as the Earth's attraction weakens. However, most of the events in our interest occur near the Earth surface, so we will get the value of gravitational acceleration as (+ 9.81) m/s^{2} when falling down and (- 9.81) m/s^{2} when moving up (as the Earth attraction in this case opposes the motion and as a result, the object slows down).

In Kinematics, Motion Maps are a useful visual tool for analysing and demonstrating what we know about how an object moves. In general, a motion map can represent the position, velocity, and acceleration of an object at various clock readings.

A motion map has many similarities with a motion graph. They both have a origin (reference point), a positive and a negative direction and they both include units. The only difference is that a motion graph doesn't show any object but only numbers. On the other hand, a motion map shows the moving object and the way how it is moving is shown through arrows. Longer the arrows, faster the object is moving.

By definition, **"Acceleration is the rate of velocity change per unit of time."**

The term "Rate" means "something changing with time." Therefore, if the change in velocity ∆v*⃗* = v*⃗*_{2} - v*⃗*_{1} occurs during a time interval ∆t, we obtain the formula of acceleration a*⃗*:

a*⃗* = *∆v**⃗**/**∆t*

=*v**⃗*_{2}-v*⃗*_{1}*/**∆t*

=

Since velocity is measured in m/s and time in s, the unit of acceleration is

Unit of acceleration = [ *m**/**s**/**s* ]= [*m**/**s*^{2} ]

Acceleration is a vector quantity as it is obtained by dividing a vector (∆v*⃗*) by a scalar (Δt).

When the velocity changes at the same rate, we have a motion with constant acceleration, otherwise, the acceleration is not constant and, in this case, it is better to consider the average acceleration.

On the other hand, the **instantaneous acceleration** shows the acceleration at a given instant, similarly to the concept of instantaneous velocity. Thus, if we are required to find the instantaneous acceleration in a certain instant t, we need to consider a very narrow time interval Δt around the given instant and calculating the respective velocities v*⃗*_{1} and v*⃗*_{2} for the two bordering points of the interval. Then, using the equation

a*⃗* = *∆v**⃗**/**∆t* = *v**⃗*_{2} - v*⃗*_{1}*/**t*_{1}-t_{1}

we find the value of the instantaneous acceleration.

If the object slows down, we say it is *decelerating*. **Deceleration** is the opposite of acceleration, i.e. it is a *"negative acceleration."* In such cases, the velocity decreases with time, i.e. the values of velocity in the successive intervals are smaller than those in the previous ones. This means the change in velocity is negative as in the expression ∆v*⃗* = v*⃗*_{i + 1} - v*⃗*_{i} (where i represents the number of the interval considered), the successive value of velocity (v*⃗*_{i + 1}) is smaller than the previous one (v*⃗*_{i}). Thus, when dividing a negative number (∆v*⃗*) by a positive one (Δt) the result is negative.

It is a known fact that Earth causes an attraction effect on objects near its surface. This attraction effect causes a downward acceleration as object move faster and faster as they approach the Earth surface. If we neglect the opposing effect of air (air resistance), we consider this acceleration as constant. It is known as the "gravitational acceleration" or the "acceleration of free fall." It is denoted by g*⃗* and has a value 9.81 m/s^{2} near the Earth surface. We get the value of gravitational acceleration as (+ 9.81) m/s^{2} when falling down and (- 9.81) m/s^{2} when moving up (as the Earth attraction in this case opposes the motion and as a result, the object slows down).

*1. The following motion map shows a moving object photographed in time clocks equal to 0.2 s each. What is the acceleration of the object if one unit in the arrows' length represents 2 m/s? *

(Hint! Consider only the arrows' length, not the distance between the dots).

- 10 m/s
^{2} - -10 m/s
^{2} - 5 m/s
^{2} - 1 m/s
^{2}

**Correct Answer: A**

*2. A car is moving at 24 m/s when immediately the driver notices an obstacle in front of him. He immediately presses the brakes and decelerates at 6m/s ^{2}. How long does it take the car to stop? *

- 96 s
- 6 s
- 4 s
- 3 s

**Correct Answer: C**

*3. The velocity vs time graph of a moving object is shown below. *

What is the instantaneous acceleration at t = 4.2 s?

- 4 m/s
^{2} - -4 m/s
^{2} - 2 m/s
^{2} - -2 m/s
^{2}

**Correct Answer: B**

We hope you found this Physics tutorial useful, if you did. Please take the time to rate this tutorial and/or share on your favourite social network. For more insight on the physical quantities of Kinematics and the relationship between them, please read Equations of Motion..

The following Physics Calculators are provided in support of the Kinematics tutorials.

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- Uniformerly Accelerated Decelerated Motion Calculator
- Projectile Motion Calculator

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