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In this article, you will learn what we mean by **instantaneous acceleration**, or more simply **acceleration**, when describing the motion of a particle.

We will see the definition and formula for instantaneous acceleration with an example that demonstrates how to use the formula in practice.

We will also cover other important things that you should know, like what the instantaneous acceleration is represented by on a **velocity vs time graph**.

The acceleration a that a particle has at an instant t is equal to the value that the average acceleration, calculated for an interval of time Î”t which includes the instant t, approaches as the interval of time Î”t gets smaller and smaller, i.e., as Î”t approaches 0.

We know that the average acceleration a for an interval of time Î”t is expressed as:

a = | Î”v |

Î”t |

where Î”v is the change in velocity that occurs during Î”t.

The formula for the instantaneous acceleration a is almost the same, except that we need to indicate that we're interested in knowing what the ratio of Î”v to Î”t approaches as Î”t approaches zero.

We can indicate that by using the limit notation.

So, the formula for the instantaneous acceleration is:

a = | lim | Î”v | |

Î”tâ†’0 | Î”t |

To demonstrate how to use this formula in practice, let's go through a simple example.

Let's consider a particle whose velocity (in meters per second) at an instant t (in seconds) is given by 2t^{2}:

v = 2t^{2}

So, at 1 s the velocity is 2 m/s, at 2 s the velocity is 8 m/s, at 3 s the velocity is 18 m/s, etc.

Let's say that we want to find the acceleration of the particle at the instant t = 3 s.

The first thing we need to do is to choose an interval of time Î”t which includes the instant 3 s.

This interval of time Î”t starts at some instant t_{1} and ends at some instant t_{2} such that

t_{1} â‰¤ 3 s â‰¤ t_{2}

For simplicity, we will choose t_{1} = 3 s so that t_{1} is as close as possible to 3 s and Î”t can be made smaller by choosing values of t_{2} that are closer to 3 s.

t_{1} = 3 s

t_{2} > 3 s

Let's begin by choosing t_{2} equal to 3.1 s.

t_{1} = 3 s

t_{2} = 3.1 s

Î”t = t_{2} âˆ’ t_{1} = 3.1 s âˆ’ 3 s = 0.1 s

The average acceleration for Î”t is equal to:

a = | Î”v | = | v_{2} âˆ’ v_{1} |

Î”t | t_{2} âˆ’ t_{1} |

Let's find the velocity v_{1} at instant t_{1}:

v_{1} = 2t_{1}^{2}

v_{1} = 2 (3)^{2} m/s

v_{1} = 18 m/s

and the velocity v_{2} at instant t_{2}:

v_{2} = 2t_{2}^{2}

v_{2} = 2 (3.1)^{2} m/s

v_{2} = 19.22 m/s

Now, we can calculate the average acceleration:

a = | v_{2} âˆ’ v_{1} |

t_{2} âˆ’ t_{1} |

a = | 19.22 m/s âˆ’ 18 m/s |

3.1 s âˆ’ 3 s |

a = | 1.22 m/s |

0.1 s |

a = 12.2 m/s^{2}

So, when the interval of time Î”t is between 3 s and 3.1 s, the average acceleration is 12.2 m/s^{2}.

Let's see what happens when we choose a smaller interval of time Î”t, with t_{2} equal to 3.01 s.

t_{1} = 3 s

t_{2} = 3.01 s

Î”t = 3.01 s âˆ’ 3 s = 0.01 s

We already know that the velocity v_{1} at instant t_{1} is 18 m/s.

The velocity v_{2} at instant t_{2} is:

v_{2} = 2t_{2}^{2}

v_{2} = 2 (3.01)^{2} m/s

v_{2} = 18.1202 m/s

The average acceleration is:

a = | v_{2} âˆ’ v_{1} |

t_{2} âˆ’ t_{1} |

a = | 18.1202 m/s âˆ’ 18 m/s |

3.01 s âˆ’ 3 s |

a = | 0.1202 m/s |

0.01 s |

a = 12.02 m/s^{2}

So, when the interval of time Î”t is between 3 s and 3.01 s, the average acceleration is 12.02 m/s^{2}.

One last time, let's choose an even smaller Î”t, with t_{2} equal to 3.001 s.

t_{1} = 3 s

t_{2} = 3.001 s

Î”t = 0.001 s

The velocity v_{2} at instant t_{2} is:

v_{2} = 2t_{2}^{2}

v_{2} = 2 (3.001)^{2} m/s

v_{2} = 18.012002 m/s

The average acceleration is:

a = | 18.012002 m/s âˆ’ 18 m/s |

3.001 s âˆ’ 3 s |

a = | 0.012002 m/s |

0.001 s |

a = 12.002 m/s^{2}

So, as Î”t is getting smaller and smaller, the average acceleration seems to be approaching 12 m/s^{2}.

We can keep choosing a smaller and smaller Î”t ad infinitum and get closer and closer to 12 m/s^{2}.

However, we can show that the average acceleration approaches 12 m/s^{2}, as Î”t gets smaller and smaller, in a more rigorous way so that we can be sure that the acceleration at the instant 3 s is 12 m/s^{2}.

First of all, we will be referring to the instant 3 s as t to indicate that what we're about to do, not only applies to the specific instant 3 s, but to any instant t.

We begin, as before, by considering an interval of time Î”t which starts at some instant t_{1} and ends at some instant t_{2} such that

t_{1} â‰¤ t â‰¤ t_{2}

Once again, we will choose t_{1} = t so that t_{1} is as close as possible to t and Î”t can be made smaller by choosing values of t_{2} that are closer to t.

t_{1} = t

t_{2} > t

Since Î”t = t_{2} âˆ’ t_{1} and t_{1} = t, we can write:

Î”t = t_{2} âˆ’ t

â†“

t_{2} = t + Î”t

This means that the boundaries of the interval of time Î”t are the instant t and the instant t + Î”t.

The velocity v_{t} at instant t is:

v_{t} = 2t^{2}

and the velocity v_{t+Î”t} at instant t + Î”t is:

v_{t+Î”t} = 2 (t + Î”t)^{2}

v_{t+Î”t} = 2 (t^{2} + 2tÎ”t + Î”t^{2})

v_{t+Î”t} = 2t^{2} + 4tÎ”t + 2Î”t^{2}

Now, we can calculate the average acceleration for Î”t:

a = | v_{t+Î”t} âˆ’ v_{t} |

Î”t |

a = | 2t^{2} + 4tÎ”t + 2Î”t^{2} âˆ’ 2t^{2} |

Î”t |

a = | 4tÎ”t + 2Î”t^{2} |

Î”t |

a = 4t + 2Î”t

Notice that this expression, 4t + 2Î”t, explains why the average acceleration that we were manually computing before was 12.2 m/s^{2} when Î”t was 0.1 s, 12.02 m/s^{2} when Î”t was 0.01 s, and 12.002 m/s^{2} when Î”t was 0.001 s:

(4 Ã— 3 + 2 Ã— 0.1) m/s^{2} = 12.2 m/s^{2}

(4 Ã— 3 + 2 Ã— 0.01) m/s^{2} = 12.02 m/s^{2}

(4 Ã— 3 + 2 Ã— 0.001) m/s^{2} = 12.002 m/s^{2}

Now, remember that the instantaneous acceleration is equal to the value that the average acceleration approaches as the interval of time Î”t approaches 0:

a = | lim | Î”v | |

Î”tâ†’0 | Î”t |

a = | lim | v_{t+Î”t} âˆ’ v_{t} | |

Î”tâ†’0 | Î”t |

In this case, we found that the average acceleration is given by 4t + 2Î”t. So,

a = | lim | 4t + 2Î”t |

Î”tâ†’0 |

As Î”t approaches 0, the term 2Î”t, within the expression 4t + 2Î”t, approaches 0, so the expression approaches 4t.

Therefore, the instantaneous acceleration a is given by 4t:

a = 4t

So, the acceleration of the particle at any instant t is given by 4t.

Thus, at the instant t = 3 s, the acceleration is 4 Ã— 3 m/s^{2}, i.e., 12 m/s^{2}.

The limit

lim | v_{t+Î”t} âˆ’ v_{t} | |

Î”tâ†’0 | Î”t |

is called the derivative of v with respect to t, which is written as

dv |

dt |

Therefore, we say that the instantaneous acceleration is the derivative of velocity with respect to time:

a = | dv |

dt |

Furthermore, since velocity is the derivative of position with respect to time:

v = | dx |

dt |

We can write the instantaneous acceleration as:

a = | dv | = | d | dx | = | d^{2}x | |

dt | dt | dt | dt^{2} |

Thus, the instantaneous acceleration is the second derivative of position with respect to time:

a = | d^{2}x |

dt^{2} |

Often, to show how the acceleration of a particle changes over time, an **acceleration vs time graph** is used.

Here's what an acceleration vs time graph might look like for a moving particle:

In this case, at the instant t = 4 s, the particle has an acceleration a = 6 m/s^{2}:

Recall that in a previous example, we found that the instantaneous acceleration of a particle was given by 4t:

a = 4t

Let's show how the acceleration of this particle changes over time with an acceleration vs time graph.

The graph of the function 4t is a line so we only need to find two points to draw it.

To keep the calculations simple, we will choose the instant t_{1} = 0 s and the instant t_{2} = 1 s:

t_{1} = 0 s, a_{1} = (4 Ã— 0) m/s^{2} = 0

t_{2} = 1 s, a_{2} = (4 Ã— 1) m/s^{2} = 4 m/s^{2}

Now that we found two points, we are ready to draw the acceleration vs time graph:

Let's consider a velocity vs time graph for the motion of a particle:

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

As we saw before, when we want to find the acceleration of the particle at an instant t, we consider an interval of time Î”t, which starts at t and ends at t + Î”t:

The acceleration at the instant t is equal to whatever the average acceleration for Î”t approaches as Î”t approaches zero.

We can use this information to determine what the instantaneous acceleration is represented by on a velocity vs time graph.

First, let's draw a secant line that passes through the points t and t + Î”t on the graph:

The slope of the secant line is equal to the average acceleration for the interval of time Î”t because Î”v/Î”t represents both the average acceleration and the slope of the secant line.

Thus, the acceleration at the instant t is equal to what the slope of the secant line approaches as Î”t approaches zero.

As Î”t gets smaller and smaller, the secant line gets closer and closer to the line tangent to the graph at the point t:

So, as Î”t approaches 0, the slope of the secant line approaches the slope of the line tangent to the graph at the point t.

Therefore, the acceleration at an instant t is equal to the slope of the line tangent to the velocity vs time graph at the point t.

If we label the slope of the tangent line with m_{T}, then we can write

a = | dv | = m_{T} |

dt |

Determining the sign of the instantaneous acceleration from a velocity vs time graph is quite simple.

To show how to do that, let's return to the same velocity vs time graph that we've seen before and let's consider an instant t at which we want to know the sign of the acceleration:

We already know that the acceleration at the instant t is equal to the slope of the line tangent to the graph at the point t.

Let's represent the tangent line and let's call Î¸ the angle that it makes with the positive t-axis:

To determine whether the acceleration at the instant t is positive, negative, or zero, we just have to look at the sign of the angle Î¸.

When the angle Î¸ is positive:

The slope of the tangent line is positive, and therefore the instantaneous acceleration is positive.

When the angle Î¸ is negative:

The slope of the tangent line is negative, and therefore the instantaneous acceleration is negative.

Finally, when the angle Î¸ is zero:

The slope of the tangent line is zero, and therefore the instantaneous acceleration is zero.

Thus, the sign of the acceleration at an instant t is the same as that of the angle Î¸ that the line, tangent to the velocity vs time graph at the point t, makes with the positive t-axis.

So, we can easily determine when the acceleration is positive, negative, and zero, just by looking at the angle Î¸ at different points on a velocity vs time graph:

When acceleration causes velocity to decrease in magnitude, it is sometimes called **deceleration**.

So, when a particle has a positive velocity and a negative acceleration or a negative velocity and a positive acceleration, you may hear the acceleration of the particle be called deceleration.