Constant acceleration

Sometimes an object speeds up or slows down at a steady rate. For example, a car joining a freeway may increase its speed by the same amount every second, or a dropped object (ignoring air resistance) gains about \(\mathbf{9.8}\,\textbf{m/s}\) of speed each second as it falls. In these situations, the acceleration stays the same with time – it is constant, or uniform.

When acceleration is constant, the velocity–time graph is a straight line. This lets us use a set of simple equations to relate displacement, velocity, acceleration and time. The same ideas apply to horizontal motion (like cars or cyclists) and vertical motion under gravity, provided the acceleration is uniform and we clearly choose which direction is positive.

Acceleration in horizontal motion

We have already used the equation \(v = \dfrac{s}{t}\) to describe motion where the velocity is constant. This equation cannot be used in situations where the velocity increases over time – that is, when acceleration occurs. Let's consider objects that accelerate at a constant rate.

You might remember that the uniform acceleration \(a\) of an object is given by the gradient of a velocity–time graph:

\[\begin{align*} a & = \frac{\text{rise}}{\text{run}} \\[6pt]
& = \frac{\text{change in velocity}}{\text{time taken}}\\[6pt]
& = \frac{\Delta v}{t}\\[6pt]
& = \frac{v_{f}-v_{i}}{t}\\[6pt]
\therefore at & = v_{f}-v_{i}\end{align*}\]

where \(v_{i}\) is the initial velocity and \(v_{f}\) is the final velocity. This can be rearranged into the form:

\[v_{f} = v_{i} + at\]

We can then use other relationships to obtain the following equations:

\[x = v_{i}t + \tfrac{1}{2}at^{2}\]

\[x = v_{f}t - \tfrac{1}{2}at^{2}\]

\[v_{f}^{2} = v_{i}^{2} + 2ax\]

\[x = \tfrac{1}{2}(v_{i} + v_{f})t\]

where \(x\) is displacement, \(v_{i}\) is initial velocity, \(v_{f}\) is final velocity and \(t\) is time taken.

Other symbols may be used to denote \(x\) (such as \(s\)), \(v_{i}\) (such as \(u\) or \(v_{0}\)) and \(v_{f}\) (such as \(v\)).

Where relevant, you must specify right or left as the positive or negative direction when doing these problems since they deal with vector quantities.

Example – using horizontal acceleration

A truck accelerates at \(\mathbf{0.8}\,\textbf{m}\,\textbf{s}\mathbf{^{-2}}\) from a starting velocity of \(\mathbf{3}\,\textbf{m}\,\textbf{s}\mathbf{^{-1}}\), until it reaches a velocity of \(\mathbf{20}\,\textbf{m}\,\textbf{s}\mathbf{^{-1}}\). Find:

1. the time taken

It is always a good idea to list the data first. We know that \(v_{i}=3\,\text{m}\,\text{s}^{-1}\), \(v_{f}=20\,\text{m}\,\text{s}^{-1}\) and \(a=0.8\,\text{m}\,\text{s}^{-2}\).

Now, we need to find the appropriate equation that includes this data, and our unknown \(t\).

\[v_{f} = v_{i} + at\]

We can either transpose the equation for \(t\) or substitute, then solve for \(t\). In this example, we will transpose first.

\[\begin{align*} t & = \frac{v_{f}-v_{i}}{a} \\[6pt]
& = \frac{20-3}{0.8}\\[6pt]
& = 21.25\,\text{s}\end{align*}\]

The smallest number of significant figures is \(1\), so we should write our answer to \(1\) significant figure. In other words, the time taken is \(20\,\text{s}\).

2. the distance travelled.

To find the displacement, we need to find a equation containing our unknown \(x\).

\[v_{f}^{2} = v_{i}^{2} + 2ax\]

Now, we can substitute the values into the rearranged equation.

\[\begin{align*} x & = \frac{v_{f}^{2}-v_{i}^{2}}{2a} \\[6pt]
& = \frac{20^{2}-3^{2}}{2\times0.8}\\[6pt]
& = \frac{400-9}{1.6}\\[6pt]
& = 244.4\,\text{m}\end{align*}\]

Acceleration due to gravity

The acceleration of a falling object near the Earth’s surface is \(9.8\,\text{m}\,\text{s}^{-2}\). For instance, a coin that is dropped from rest will have a velocity of \(9.8\,\text{m}\,\text{s}^{-1}\) after \(1\,\text{s}\), \(19.6\,\text{m}\,\text{s}^{-1}\) after \(2\,\text{s}\), and so on. Acceleration due to gravity can also be expressed in units of \(\text{g}\), where \(1\,\text{g}=9.8\,\text{m}\,\text{s}^{-2}\).

Since the acceleration of a freely falling object is uniform (constant), we can use the same equations as we did with horizontal motion. Similarly, we may need to specify up or down as the positive or negative direction as acceleration is a vector quantity.

Example 1 – using vertical acceleration

A construction worker accidentally knocks a brick from a building so that it falls vertically a distance of \(\mathbf{50}\,\textbf{m}\) to the ground. Calculate:

1. the time taken for the brick to reach the ground

List the data first: \(v_{i}=0\), \(x=50\,\text{m}\) and \(a=9.8\,\text{m}\,\text{s}^{-2}\).

We need to find \(t\), so the equation we need is:

\[x = v_{i}t+\tfrac{1}{2}at^{2}\]

Here, we will substitute first, then solve for \(t\).

\[\begin{align*} 50 & = \left(0\times t\right)+\left(\frac{1}{2}\times9.8\times t^{2}\right) \\[6pt]
50 & = 0+4.9t^{2}\\[6pt]
\frac{50}{4.9} & = t^{2}\\[6pt]
10.2 & = t^{2}\\[6pt]
\sqrt{10.2} & = t\\[6pt]
t & = 3.2\,\text{s}\end{align*}\]

2. the speed of the brick as it hits the ground.

The speed of the brick when it hits the ground is the final velocity or \(v_{f}\). The relevant equation is:

\[v_{f}=v_{i}+at\]

Next, we take our data and substitute to find \(v_{f}\).

\[\begin{align*} v_{f} & = 0+\left(9.8\times3.2\right) \\[6pt]
& = 31.3\,\text{m}\,\text{s}^{-1}\end{align*}\]

Exercise – using acceleration

For the following questions, use \(g=9.8\,\text{m}\,\text{s}^{-2}\).

1. A car, starting from rest, moves with an acceleration of \(2\,\text{m}\,\text{s}^{-2}\). Find:
   1. the velocity at the end of \(20\,\text{s}\)
   2. the distance covered in that time.

2. Determine the uniform acceleration required for a spacecraft, starting from rest, to cover \(1000\,\text{m}\) in \(10\,\text{s}\).

3. A cyclist, starting from rest, moves with an acceleration of \(3\text{m}\,\text{s}^{-2}\). Find:
   1. the time taken to reach a velocity of \(30\,\text{m/s}\)
   2. the distance the cyclist covers in this time.

4. An object starts with a velocity of \(100\,\text{m/s}\) and decelerates (slows down) at \(2\text{m}\,\text{s}^{-2}\). Find:
   1. at what time its velocity will be zero
   2. how far it will have gone.

5. A book is knocked off a ledge and falls vertically to the floor. If the book takes \(1.0\,\text{s}\) to fall to the floor, calculate:
   1. its speed as it lands
   2. the height from which it fell
   3. the distance it falls during the first \(0.5\,\text{s}\)
   4. the distance it falls during the final \(0.5\,\text{s}\).

6. A champagne cork travels vertically into the air. It takes \(4.0\,\text{s}\) to return to its starting position. Determine:
   1. how long the cork takes to reach its maximum height
   2. the maximum height reached by the cork
   3. how fast the cork was travelling initially
   4. the speed of the cork as it returned to its starting point
   5. the acceleration of the cork at each of these times after its launch:
      1. \(1.0\,\text{s}\)
      2. \(2.0\,\text{s}\)
      3. \(3.0\,\text{s}\).

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