# Non-Uniform Acceleration

## Non-Uniform Acceleration Revision

**Non-Uniform Acceleration**

When the acceleration of a particle isn’t constant, we have to ignore the SUVAT equations. We can use calculus to relate **Displacement**, **Velocity** and **Acceleration**.

Make sure you are happy with the following topics before continuing.

**Relationships between \textbf{s}, \textbf{v} and \textbf{a}**

Think about what velocity and acceleration actually mean for a moment.

**Velocity**is a change in displacement, over a period of time.**Acceleration**is a change in velocity, over a period of time.

Think back to our **Displacement-Time** graphs. We said that the gradient of the DT graph gives the **velocity** at any moment in time.

What about the **Velocity-Time** graphs? We said that the gradient represented the **acceleration**, and that the area under the graph was the **distance** travelled.

Well, we know how to find a gradient of a function at any point – that’s **differentiation**.

And we also know how to find the area under a function between two points – that’s **integration**.

That gives us a simple flowchart:

Remember, we’ve got to be careful with integration. Forgetting the + c term can throw our valuations of \textcolor{red}{v} and \textcolor{limegreen}{s} off wildly.

**Finding Maxima and Minima**

This calculus operates in the same way that regular calculus does.

So, for example, let’s say we want to find the maximum or minimum **velocity** at any given point. Well, logically, we’re looking for a point in time where there’s no **acceleration** or **deceleration**. You might say, when the **change in velocity**, \dfrac{d\textcolor{red}{v}}{d\textcolor{purple}{t}} = 0.

Then, we’d differentiate this again to find out whether this “stationary point” is a minimum, maximum or point of inflection.

The same goes for finding a maximum or minimum **displacement** from a starting point, too.

We’re looking for a point where the **velocity**, \textcolor{red}{v} = \dfrac{d\textcolor{limegreen}{s}}{d\textcolor{purple}{t}}, is 0.

Beyond there, we’re looking to find out whether our point is a minimum, maximum or point of inflection.

So, to recap, a maximum/minimum **displacement** can be found when \textcolor{red}{v} = 0, and we can determine its nature by finding out the value of \textcolor{blue}{a}.

**Note**

The second derivative being equal to 0 does not always mean there is a point of inflection. How to handle this result is covered on the **stationary points** page.

**Deriving the SUVAT Equations**

We can use our new equations to give our SUVAT equations.

Say we have \textcolor{blue}{a} = \dfrac{d\textcolor{red}{v}}{d\textcolor{purple}{t}}

Then, \textcolor{red}{v} = \int \textcolor{blue}{a}\, d\textcolor{purple}{t} = \textcolor{blue}{a}\textcolor{purple}{t} + c

Remember, we can find c by setting \textcolor{purple}{t} = 0. Well, we label our initial velocity as u, so we can deduce that c = u, and by extension,

\textcolor{red}{v} = u + \textcolor{blue}{a}\textcolor{purple}{t}

Also, \textcolor{red}{v} = \dfrac{d\textcolor{limegreen}{s}}{d\textcolor{purple}{t}}

Then we have \textcolor{limegreen}{s} = \int \textcolor{red}{v} d\textcolor{purple}{t} = \int (u + \textcolor{blue}{a}\textcolor{purple}{t})\, d\textcolor{purple}{t} = u\textcolor{purple}{t} + \dfrac{1}{2}\textcolor{blue}{a}\textcolor{purple}{t}^2 + c

Now, we didn’t talk about what we label c when \textcolor{purple}{t} = 0. Nothing to worry about though, we’ll just call it s_0, the initial displacement point. So, we have

\textcolor{limegreen}{s} = u\textcolor{purple}{t} + \dfrac{1}{2}\textcolor{blue}{a}\textcolor{purple}{t}^2 + s_0

**Example 1: Finding a Minimum or Maximum Displacement**

Let’s put this into action.

Say a particle has an equation for displacement, \textcolor{limegreen}{s} = 3\textcolor{purple}{t}^3 - \textcolor{purple}{t} - 9.

Find the point where the particle is stationary, and find the nature of this stationary point.

**[3 marks]**

\dfrac{d\textcolor{limegreen}{s}}{dt} = \textcolor{red}{v} = 9\textcolor{purple}{t}^2 - 1 = (3\textcolor{purple}{t} + 1)(3\textcolor{purple}{t} - 1)\\

\textcolor{red}{v} = \textcolor{red}{0} when \textcolor{purple}{t} = Â±\textcolor{purple}{\dfrac{1}{3}}

Of course, we can’t have a negative time, so we only have \textcolor{purple}{t} = \textcolor{purple}{\dfrac{1}{3}}.

\dfrac{d^{2}\textcolor{limegreen}{s}}{d\textcolor{purple}{t}^{2}} = \dfrac{d\textcolor{red}{v}}{d\textcolor{purple}{t}} = \textcolor{blue}{a} = 18\textcolor{purple}{t}

When \textcolor{purple}{t} = \textcolor{purple}{\dfrac{1}{3}}, \textcolor{blue}{a} = \textcolor{blue}{6}\\ 6 > 0, so we conclude that the **displacement** is at a **minimum** when \textcolor{purple}{t} = \textcolor{purple}{\dfrac{1}{3}}\text{ s}.

**Example 2:Â Finding a Minimum or Maximum Velocity**

Now, say a particle has a **velocity** of \textcolor{red}{v} = 25 + 8\textcolor{purple}{t} - \textcolor{purple}{t}^2.

Find the time \textcolor{purple}{t} when the particle is not **accelerating** or **decelerating**. Find out whether this point corresponds to the minimum or maximum **velocity**.

**[3 marks]**

\dfrac{d\textcolor{red}{v}}{d\textcolor{purple}{t}} = \textcolor{blue}{a} = 8 - 2\textcolor{purple}{t}

We have \textcolor{blue}{a} = \textcolor{blue}{0} when \textcolor{purple}{t} = \textcolor{purple}{4}.

\dfrac{d^{2}\textcolor{red}{v}}{d\textcolor{purple}{t}^{2}} = \dfrac{d\textcolor{blue}{a}}{d\textcolor{purple}{t}} = \textcolor{blue}{-2}

Since -2 < 0, we can conclude that we have a **maximum velocity** when \textcolor{purple}{t} = 4\text{ s}.

**Example 3: Finding Displacement from Acceleration**

A particle is accelerating at 6\textcolor{purple}{t}\text{ ms}^{-2}. Given that the initial velocity, u, is 10\text{ ms}^{-1}, and the initial displacement, s_0 is 50\text{ m}, find an equation for \textcolor{limegreen}{s} in terms of \textcolor{purple}{t}.

**[4 marks]**

From \textcolor{blue}{a}= 6\textcolor{purple}{t}, we have

\textcolor{red}{v} = \int \textcolor{blue}{a}\, d\textcolor{purple}{t} = \int 6\textcolor{purple}{t}\, d\textcolor{purple}{t}\\ = 3\textcolor{purple}{t}^2 + u = 3\textcolor{purple}{t}^2 + 10

By extension,

\textcolor{limegreen}{s} = \int \textcolor{red}{v}\, d\textcolor{purple}{t} = \int (3\textcolor{purple}{t}^2 + 10)\, d\textcolor{purple}{t}\\ = \textcolor{purple}{t}^3 + 10\textcolor{purple}{t} + s_0 = \textcolor{purple}{t}^3 + 10\textcolor{purple}{t} + 50

## Non-Uniform Acceleration Example Questions

**Question 1:** For a particle with displacement s = \dfrac{1}{2}t^2 - 16\sqrt{t}, find the time t where the particle is not moving, and determine whether it is accelerating or decelerating at this point.

**[3 marks]**

s = \dfrac{1}{2}t^2 - 16\sqrt{t}

gives

v = \dfrac{ds}{dt} = t - \dfrac{8}{\sqrt{t}}

Then v = 0 when t = 4. Also,

a = \dfrac{dv}{dt} = \dfrac{d^{2}s}{dt^2} = 1 + \dfrac{4}{t\sqrt{t}}

Substituting t = 4, we can see that a = 1 + \dfrac{4}{8} = 1.5, so the particle is accelerating from a minimum point.

**Question 2:** A car is accelerating at a rate of -2t\text{ ms}^{-2}, with an initial velocity of 25\text{ ms}^{-1}. Assuming it does not start travelling backwards when its velocity is 0\text{ ms}^{-1}, how far does the car travel?

**[3 marks]**

a = -2t\\

gives

v = \int a\, dt = \int (-2t)\, dt = u - t^2 = 25 - t^2

meaning v = 0 when t = 5.

s = \int v\, dt = \int (25 - t^2)\, dt = 25t - \dfrac{1}{3}t^3 + c = 25t - \dfrac{1}{3}t^3

When t = 5, s = 125 - \left( \dfrac{1}{3} \times 125\right) = 83.33\text{ m (to } 2 \text{ dp)}

**Question 3:** A particle sets off from the origin point, and its displacement is described by s = \sin t + \cos t - 1, where t is measured in radians. Find an expression for its acceleration at time t. List all times 0 \leq t \leq 2\pi where the particle is stationary.

**[4 marks]**

s = \sin t + \cos t - 1

gives

v = \cos t - \sin t

and

a = - \sin t - \cos t

We know that v = 0 when \cos t = \sin t.

So we have v = 0 when \tan t = 1.

This means that the particle is stationary when t = \dfrac{\pi}{4} and \dfrac{5\pi}{4}.

## You May Also Like...

### MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic! No fees, no trial period, just totally free access to the UKâ€™s best GCSE maths revision platform.