Instantaneous acceleration: Definition, Formula, Examples

On the basis of the time interval considered for calculation, the acceleration is classified as average acceleration and instantaneous acceleration. In this article, we are discussing instantaneous acceleration in detail.

What is instantaneous acceleration?

Instantaneous acceleration is the acceleration of the object at a specific instant during the motion. It indicates the change in velocity per unit time measured for a very small interval ‘dt’.

The object moving in a straight line may undergo an increase or decrease in acceleration or it may move with a uniform acceleration or zero acceleration. Thus in such cases, the average acceleration does not describe the motion of the object at every instant.

What is instantaneous acceleration

The average acceleration only provides the mean value of the acceleration instead of the actual acceleration of the object during the motion. While the instantaneous acceleration gives the exact acceleration at every instant during the motion.

In the above acceleration-time graph, the `a_{\text{average}}` indicates the average acceleration of the object throughout the motion while the `a_{1}` and `a_{2}` indicates the instantaneous acceleration of the object at specific instant `t_{1}` and `t_{2}` respectively.

Equations:

In the form of limits, the instantaneous acceleration of the object is given by,

`a_{\text{instantaneous}}` = `\lim_{\Delta t \to 0} \frac{V_{(t + \Delta t)}-V_{t}}{\Delta t}`

In the form of differentiation, the instantaneous acceleration of the object is given by,

`a_{\text{instantaneous}}` = `\frac{dV}{dt}`

How to find the instantaneous acceleration?

It can be found by use of the following two methods:-

1] Analytical method:

This method is used to find the instantaneous acceleration when the equation of velocity in terms of time is given. It can be solved by using the method of limits or by using differentiation.

By using limits, the instantaneous acceleration is calculated as,

`a_{\text{instantaneous}}` = `\lim_{\Delta t \to 0} \frac{V_{(t + \Delta t)}-V_{t}}{\Delta t}`

By using differentiation, the instantaneous velocity can be calculated as,

`a_{\text{instantaneous}}` = `\frac{dV}{dt}`

2] Graphical method:-

This method is used for calculating instantaneous acceleration from the velocity-time graph.

How to find instantaneous acceleration from a velocity-time graph?

The instantaneous acceleration from the different profiles of a velocity-time graph is calculated as follows:-

Case 1] When a velocity-time graph is linear:-

If the velocity-time plot has a linear nature, then it means that the object is moving with constant acceleration throughout its motion.

Instantaneous acceleration from linear velocity-time graph

Thus in this case, the instantaneous acceleration is equal to the average acceleration of the object.

`V_{\text{Instantaneous}}` = `V_{\text{Average}}`

`V_{\text{Instantaneous}}` = `\frac{\Delta V}{\Delta t}`

`V_{\text{Instantaneous}}` = `\frac{V_{2}-V_{1}}{t_{2}-t_{1}}`

Case 2] When velocity-time profile is non-linear:-

If the velocity time profile is non-linear thus it means that the acceleration of the objeat is changing throughout the motion.

Instantaneous acceleration from the graph

Thus in this case, the instantaneous acceleration of an object is different from the average acceleration. From such a graph, the instantaneous acceleration can be calculated by the use of the following steps:-

Step 1] Locate a point of a curve at a required time:-

Locate the point on the graph at instant t

Step 2] At this point draw a tangent to the curve:-

Draw a tangent at point a

Step 3] Find the slope of the tangent:- The instantaneous acceleration is the slope of the tangent drawn to the velocity-time curve at that instant.

Find the slope of the tangent

∴ The instantaneous acceleration is given by,

`a_{\text{Instantaneous}}` = Slope(1-2) = `\frac{V_{2}-V_{1}}{t_{2}-t_{1}}`

Average acceleration vs Instantaneous acceleration:

Sr. No.Average accelerationInstantaneous acceleration
1Average acceleration gives the change in velocity within a considerable duration.It gives the acceleration of the object at a particular instant.
2It is calculated for a considerable time interval ‘`\Deltat`’.It is calculated for the smallest interval of time ‘dt’.
3It gives the average value of acceleration during a specified interval.It gives exact acceleration at any instant.
4It is given by,
`a_{\text{average}}` = `\frac{\DeltaV}{\Delta t}`
It is given by,
`a_{\text{instantaneous}}` = `\frac{dV}{dt}`

Solved examples:

1] The particle is moving in a straight line with the velocity of `(5t^{2} +3t)` m/s. What is the instantaneous acceleration of the particle at t = 45.0s?

Given:
V = (`5t^{2}` +3t) m/s
t = 45 seconds

Solution:-

Method 1:- Using limits

The instantaneous acceleration of the object is given by,

`a_{t}` = `\lim_{\Delta t \to 0} \frac{V_{(t + \Delta t)}-V_{t}}{\Delta t}`

`a_{t}` = `\lim_{\Delta t \to 0} \frac{[5(t + \Delta t)^{2} + 3(t + \Delta t)] – [5t^{2} +3t]}{\Delta t}`

`a_{t}` = `\lim_{\Delta t \to 0} \frac{5t^{2} + 10t\Delta t + 5\Delta t^{2}+3t+3\Delta t-5t^{2}-3t}{\Delta t}`

`a_{t}` = `\lim_{\Delta t \to 0} \frac{5\Delta t^{2}+10t\Delta t+3\Delta t}{\Delta t}`

`a_{t}` = `\lim_{\Delta t \to 0} 5\Delta t` + 10t + 3

`a_{t}` = 10t + 3

Now the acceleration of the object at t = 45 seconds is given by,

`a_{(t=45)}` = 10(45) + 3

`a_{(t=45)}` = 453 m/s²
Method 2:- Using differentiation

`a_{t}` = `\frac{dV}{dt}`

`a_{t}` = `\frac{d}{dt}(5t^{2} + 3t)`

`a_{t}` = 10t + 3

The acceleration at t = 45 seconds is given by,

`a_{(t=45)}` = 10(45) + 3

`a_{(t=45)}` = 453 m/s²

2] From the given velocity-time plot, find the acceleration of the object at t = 3.5 seconds.
instantaneous acceleration numerical

Given:
t = 3.5 seconds

Solution:-

Step 1] Locate point on curve at time t = 3.5 seconds:

Locate the point

Step 2] Draw a tangent to the curve at point x:

tangent at marked point

Step 3] Find the slope of the tangent:

The slope of the tangent

∴ The instantaneous acceleration at t = 3.5 second is given by,

`a_{(t=3.5)}` = Slope(1-2) = `\frac{V_{2}-V_{1}}{t_{2}-t_{1}}`

From above figure,

`V_{2}` = 0.5 m/s, `V_{1}`= 3 m/s,
`t_{2}` = 6 seconds, `t_{1}` = 1 second.

`a_{(t=3.5)}` = `\frac{0.5-3}{6-1}`

`a_{(t=3.5)}` = – 0.5 m/s²

Thus at t = 3.5 seconds, the object has a retardation of 0.5 m/s².

3] A particle starting from rest is moving at the velocity of (6t² + 2t) m/s. If the average acceleration of the particle is 6 m/s², then find the time when average acceleration is equal to instantaneous acceleration.

Given:
V = (`6t^{2}` + 2t) m/s
`a_{\text{average}}` = 6 m/s²

Solution:-

The instantaneous acceleration of the particle is given by,

`a_{\text{instantaneous}}` = `\frac{dV}{dt}`

`a_{\text{instantaneous}}` = `\frac{d}{dt}(6t^{2} + 2t)`

`a_{\text{instantaneous}}` = 12t + 2

The time at which the `V_{\text{average}}` is equals to the `V_{\text{instantaneous}}` is given by,

`a_{\text{average}}` = `a_{\text{instantaneous}}`

6 = 12t + 2

t = 0.34 seconds

FAQs:

  1. Why instantaneous acceleration is important?

    It gives the acceleration of the object at a specific instant during the motion.

  2. Is instantaneous acceleration always changing?

    For the object moving with constant acceleration or constant retardation, the instantaneous acceleration never changes while for the object moving with continuously changing its acceleration, the instantaneous acceleration of the object always changes.

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