21.7 Equations of motion

21.7 Equations of motion (ESAHG)

In this section we will look at the third way to describe motion. We have looked at describing motion in terms of words and graphs. In this section we examine equations that can be used to describe motion.

This section is about solving problems relating to uniformly accelerated motion. In other words, motion at constant acceleration.

The following are the variables that will be used in this section:

An alternate convention for some of the variables exists that you will likely encounter so here is a list for reference purposes:

In this book we will use the first convention.

The questions can vary a lot, but the following method for answering them will always work. Use this when attempting a question that involves motion with constant acceleration. You need any three known quantities (\({\vec{v}}_{i}\), \({\vec{v}}_{f}\), \(\Delta \vec{x}\), \(t\) or \(\vec{a}\)) to be able to calculate the fourth one.

Problem solving strategy:

1. Read the question carefully to identify the quantities that are given. Write them down.

2. Identify the equation to use. Write it down!!!

3. Ensure that all the values are in the correct units and fill them in your equation.

4. Calculate the answer and check your units.

Extension: Finding the equations of motion (ESAHH)

The following does not form part of the syllabus and can be considered additional information.

Derivation of Equation 1

According to the definition of acceleration:

\[\vec{a}=\frac{\Delta \vec{v}}{t}\]

where \(\Delta \vec{v}\) is the change in velocity, i.e. \(\Delta v={\vec{v}}_{f} - {\vec{v}}_{i}\). Thus we have

\begin{align*} \vec{a} & = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{t} \\ {\vec{v}}_{f} & = {\vec{v}}_{i} + \vec{a}t \end{align*}

Derivation of Equation 2

We have seen that displacement can be calculated from the area under a velocity vs. time graph. For uniformly accelerated motion the most complicated velocity vs. time graph we can have is a straight line. Look at the graph below - it represents an object with a starting velocity of \({\vec{v}}_{i}\), accelerating to a final velocity \({\vec{v}}_{f}\) over a total time t.

To calculate the final displacement we must calculate the area under the graph - this is just the area of the rectangle added to the area of the triangle. This portion of the graph has been shaded for clarity.

\begin{align*} {\text{Area}}_{△} & = \frac{1}{2} b\times h \\ & = \frac{1}{2} t \times \left({v}_{f} - {v}_{i}\right) \\ & = \frac{1}{2}{v}_{f}t - \frac{1}{2}{v}_{i}t \end{align*}\begin{align*} {\text{Area}}_{\square } & = l \times b \\ & = t\times {v}_{i} \\ & = {v}_{i}t \end{align*}\begin{align*} \text{Displacement } & = {\text{Area}}_{\square } + {\text{Area}}_{△} \\ \Delta \vec{x} & = {v}_{i}t + \frac{1}{2}{v}_{f}t - \frac{1}{2}{v}_{i}t \\ \Delta \vec{x} & = \frac{\left({v}_{i} + {v}_{f}\right)}{2}t \end{align*}

Derivation of Equation 3

This equation is simply derived by eliminating the final velocity \({v}_{f}\) in Equation 2. Remembering from Equation 1 that

\[{\vec{v}}_{f} = {\vec{v}}_{i} + \vec{a}t\]

then Equation 2 becomes

\begin{align*} \Delta \vec{x} & = \frac{{\vec{v}}_{i} + {\vec{v}}_{i} + \vec{a}t}{2}t \\ & = \frac{2{\vec{v}}_{i}t + \vec{a}{t}^{2}}{2} \\ \Delta \vec{x} & = {\vec{v}}_{i}t + \frac{1}{2}\vec{a}{t}^{2} \end{align*}

Derivation of Equation 4

This equation is just derived by eliminating the time variable in the above equation. From Equation 1 we know

\[t = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{a}\]

Substituting this into Equation 3 gives

\begin{align*} \Delta \vec{x} & = {\vec{v}}_{i}\left(\frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{a}\right) + \frac{1}{2}a{\left(\frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{\vec{a}}\right)}^{2} \\ & = \frac{{\vec{v}}_{i}{\vec{v}}_{f}}{\vec{a}} - \frac{{\vec{v}}_{i}^{2}}{\vec{a}} + \frac{1}{2}\vec{a}\left(\frac{{\vec{v}}_{f}^{2} - 2{\vec{v}}_{i}{\vec{v}}_{f} + {\vec{v}}_{i}^{2}}{{\vec{a}}^{2}}\right) \\ & = \frac{{\vec{v}}_{i}{\vec{v}}_{f}}{\vec{a}} - \frac{{\vec{v}}_{i}^{2}}{\vec{a}} + \frac{{\vec{v}}_{f}^{2}}{2\vec{a}} - \frac{{\vec{v}}_{i}{\vec{v}}_{f}}{\vec{a}} + \frac{{\vec{v}}_{i}^{2}}{2\vec{a}} \\ 2\vec{a}\Delta \vec{x} & = -2{\vec{v}}_{i}^{2} + {\vec{v}}_{f}^{2} + {\vec{v}}_{i}^{2} \\ {\vec{v}}_{f}^{2} & = {\vec{v}}_{i}^{2} + 2\vec{a}\Delta \vec{x} \end{align*}

This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable.

Applications in the real-world (ESAHI)

What we have learnt in this chapter can be directly applied to road safety. We can analyse the relationship between speed and stopping distance. The following worked example illustrates this application.

Worked example 9: Stopping distance

A truck is travelling at a constant velocity of \(\text{10}\) \(\text{m·s$^{-1}$}\) when the driver sees a child \(\text{50}\) \(\text{m}\) in front of him in the road. He hits the brakes to stop the truck. The truck accelerates at a rate of \(-\text{1,25}\) \(\text{m·s$^{-2}$}\). His reaction time to hit the brakes is \(\text{0,5}\) seconds. Will the truck hit the child?

Analyse the problem and identify what information is given

It is useful to draw a time-line like this one:

We need to know the following:

• What distance the driver covers before hitting the brakes.

• How long it takes the truck to stop after hitting the brakes.

• What total distance the truck covers to stop.

Calculate the distance \(AB\)

Before the driver hits the brakes, the truck is travelling at constant velocity. There is no acceleration and therefore the equations of motion are not used. To find the distance travelled, we use:

\begin{align*} v & = \frac{D}{t} \\ 10 & = \frac{D}{\text{0,5}} \\ D & = \text{5}\text{ m} \end{align*}

The truck covers \(\text{5}\) \(\text{m}\) before the driver hits the brakes.

Calculate the time \(BC\)

We have the following for the motion between \(B\) and \(C\):

\begin{align*} {\vec{v}}_{i} & = \text{10}\text{ m·s$^{-1}$} \\ {\vec{v}}_{f} & = \text{0}\text{ m·s$^{-1}$} \\ a & = -\text{1,25}\text{ m·s$^{-2}$} \\ t & = ? \end{align*}

We can use Equation 1:

\begin{align*} {\vec{v}}_{f} & = {\vec{v}}_{i} + at \\ 0 & = \text{10}\text{ m·s$^{-1}$} + \left(-\text{1,25}\text{ m·s$^{-2}$}\right)t \\ -\text{10}\text{ m·s$^{-1}$} & = \left(-\text{1,25}\text{ m·s$^{-2}$}\right)t \\ t & = \text{8}\text{ s} \end{align*}

Calculate the distance \(BC\)

For the distance we can use Equation 2 or Equation 3. We will use Equation 2:

\begin{align*} \Delta \vec{x} & = \frac{\left({\vec{v}}_{i} + {\vec{v}}_{f}\right)}{2}t \\ \Delta \vec{x} & = \frac{10 + 0}{2} \left(8\right) \\ \Delta \vec{x} & = \text{40}\text{ m} \end{align*}

Write the final answer

The total distance that the truck covers is \({d}_{AB} + {d}_{BC} = \text{5} + \text{40} = \text{45}\) meters. The child is \(\text{50}\) meters ahead. The truck will not hit the child.