Dynamic Effects of linear Motion

Variable Acceleration Depending on Time

In Engineering, motion is often analysed through quantities like displacement, velocity, and acceleration. While many introductory problems assume constant acceleration (like free-fall near Earth’s surface), real-world motion is frequently more complex. One important case is when acceleration changes with time, this is known as variable acceleration depending on time.

Variable acceleration means that the rate at which an object’s velocity changes is not fixed, but instead varies as time progresses. This can happen in many scenarios: a car gradually pressing the gas pedal harder, a rocket burning fuel at a decreasing rate, or a ball slowing down due to changing friction.

To analyse such motion, we often express acceleration as a function of time, written as a(t). From this function, we can find the velocity v(t) and position x(t) of the object using integration, because acceleration is the derivative of velocity, and velocity is the derivative of displacement.

Understanding variable acceleration is essential for describing and predicting motion in dynamic environments where forces change over time. It lays the groundwork for more advanced concepts in mechanics and helps bridge the gap between simple models and real-life behaviour.

Note that ‘position’ (relative to a datum) and ‘displacement’ are used interchangeably in this article. 

Recall, that acceleration, a, velocity, v and displacement, x are related by:

a = ^dv⁄_dt

v = ^dx⁄_dt

a = ^d²x⁄_dt²

It follows then that:

v = ∫ a dt

And 

x = ∫ v dt

If acceleration is a function of time:

a = f(t)

So that:

v = ∫ f(t) dt + c

Where c is an arbitrary constant of integration (see Engineering Mathematics unit for a recap on integration if this is unfamiliar).

Similarly, if the velocity is a function of time, t, i.e.

a = g(t)

Then in the same manner as above:

x = ∫ g(t) dt + c

Example 1

A vehicle is travelling along a straight road and is known to move with an acceleration given by the function ²⁄₃t m/s²

 When the time, t = 3 seconds the speed of the vehicle is measured to be 4 m/s and its position, x  from the starting position (point 0) is 10m. Find the position (displacement) of the vehicle when time, t = 6 seconds.

Let us start with writing down the equation for the acceleration, a:

a = ^dv⁄_dt

We know that the acceleration is given by the function ²⁄₃t m/s² and so we can state:

a = ^dv⁄_dt = ²⁄₃t

Let us now find the velocity, by integrating:

v = ∫ a dt

v = ∫ ²⁄₃ t dt

v =  ²⁄₃ × ^t2⁄₂ + c

v =  ²⁄₃ × (t²) + c

where c is some arbitrary constant of integration.

We know from the information we are given that: v = 4 m/s when t = 3 seconds. So let us substitute these values into the above equation for v, to find the value of c.

4 = ¹⁄₃ × 3² + c

4 = ¹⁄₃ × 9 + c

4 = 3 + c

c = 1

A piston moves in a cylinder with acceleration of a(t) = cos(πt) m/s² at time, t, seconds. The piston starts from rest at t = 0. Find the distance travelled in the time interval from t =1 to t = 2 seconds.

Acceleration, a, is given by:

a = ^dv⁄_dt = cos(πt)

The velocity, v, is given by:

 v = ∫ a dt

v = cos(πt) dt

v = ¹⁄_π sin(πt) +c

where c is some arbitrary constant of integration.

The question tells us that the piston ‘starts from rest’, and this tells us that v = 0 when t = 0. We can substitute these values into the above equation to find the value of c:

0 = ¹⁄_π sin⁡(0) + c

0 = 0 + c

c = 0

Therefore,

v = ¹⁄_π sin(πt)

Now consider displacement, x:

x = ∫ v dt

x = ∫  ¹⁄_π  sin(πt) dt

x = –¹⁄_π²  cos(πt) + c

where c is some arbitrary constant of integration. 

In this case, we have no information about the displacement at any instant in time, however, we will see that this does not matter…

Find the displacement, x, when t = 1:

x₁ =^-1⁄_π²  cos(πt) + c

x₁ = ^-1⁄_π²  cos(π) + c

Since

cosπ = -1

So

x₁ = ^-1⁄_π²  (-1) + c 

x₁ = ¹⁄_π² + c

Now, find the displacement, x, when t = 2:–

x₂ = ^-1⁄_π²  cos(2π) + c 

Since

cos2π =1

So

x₂ = ^-1⁄_π² + c 

Finally, the distance travelled between t = 1 and t = 2 seconds is given by:

|x₂ – x₁| = |(^-1⁄_π²+ c) – (¹⁄_π²+ c) |

|x₂ – x₁| = ²⁄_π²

As you can see, the constant of integration, c, cancels out and the distance moved by the piston in the given time frame is ²⁄_π² metres.

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