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Welcome to our Physics lesson on **Constant and non-constant acceleration**, this is the third lesson of our suite of physics lessons covering the topic of **The Meaning of Acceleration. Constant and Non-Constant Acceleration. Gravitational Acceleration**, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

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.

You have reach the end of Physics lesson **3.7.3 Constant and non-constant acceleration**. There are 5 lessons in this physics tutorial covering **The Meaning of Acceleration. Constant and Non-Constant Acceleration. Gravitational Acceleration**, you can access all the lessons from this tutorial below.

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