Equations of Motion Examples

A racing car is travelling North. It accelerates uniformly covering a distance of \(\text{725}\) \(\text{m}\) in \(\text{10}\) \(\text{s}\). If it has an initial velocity of \(\text{10}\) \(\text{m·s$^{-1}$}\), find its acceleration.

Step 1: Identify what information is given and what is asked for

We are given:

\begin{align*} {\vec{v}}_{i} & = \text{10}\text{ m·s$^{-1}$} \\ \Delta \vec{x} & = \text{725}\text{ m} \\ t & = \text{10}\text{ s} \\ \vec{a} & = ? \end{align*}

Step 2: Find an equation of motion relating the given information to the acceleration

If you struggle to find the correct equation, find the quantity that is not given and then look for an equation that has this quantity in it.

We can use

\[\Delta \vec{x} = {\vec{v}}_{i}t + \frac{1}{2}\vec{a}{t}^{2}\]

Step 3: Substitute your values in and find the answer

\begin{align*} \Delta \vec{x} & = {\vec{v}}_{i}t + \frac{1}{2}\vec{a}{t}^{2} \\ \text{725}\text{ m} & = \left(\text{10}\text{ m·s$^{-1}$} \times \text{10}\text{ s} \right) + \frac{1}{2} \vec{a} \times \left(\text{10}\text{ s}\right)^{2} \\ \text{725}\text{ m} - \text{100}\text{ m} & = \left(\text{50}\text{ s$^{2}$}\right)\vec{a} \\ \vec{a} & = \text{12,5}\text{ m·s$^{-2}$} \end{align*}

Step 4: Quote the final answer

The racing car is accelerating at \(\text{12.5}\) \(\text{m·s$^{-2}$}\) North.

A motorcycle, travelling East, starts from rest, moves in a straight line with a constant acceleration and covers a distance of \(\text{64}\) \(\text{m}\) in \(\text{4}\) \(\text{s}\). Calculate

• its acceleration

• its final velocity

• at what time the motorcycle had covered half the total distance

• what distance the motorcycle had covered in half the total time.

Step 1: Identify what information is given and what is asked for

We are given:

\begin{align*} \vec{v}_{i} & = \text{0}\text{ m·s$^{-1}$} \text{ (because the object starts from rest)} \\ \Delta \vec{x} & = \text{64}\text{ m} \\ t & = \text{4}\text{ s} \\ \vec{a} & = ? \\ \vec{v}_{f} & = ? \\ t & = ? \text{ at half the distance } \Delta \vec{x} = \text{32}\text{ m} \\ \Delta \vec{x} & = ? \text{ at half the time } t \text{2}\text{ s} \end{align*}

All quantities are in SI units.

Step 2: Acceleration: Find a suitable equation to calculate the acceleration

We can use

\[\Delta \vec{x}={\vec{v}}_{i}t+\frac{1}{2}\vec{a}{t}^{2}\]

Step 3: Substitute the values and calculate the acceleration

\begin{align*} \Delta \vec{x} & = {\vec{v}}_{i}t+\frac{1}{2}\vec{a}{t}^{2} \\ \text{64}\text{ m} & = \left(\text{0}\text{ m·s$^{-1}$} \times \text{4}\text{ s}\right) + \frac{1}{2}\vec{a} \times (\text{4}\text{ s})^{2} \\ \text{64}\text{ m} & = \left(\text{8}\text{ s$^{2}$}\right) \vec{a} \\ \vec{a} & = \text{8}\text{ m·s$^{-2}$} \text{ East} \end{align*}

Step 4: Final velocity: Find a suitable equation to calculate the final velocity

We can use Equation 1 - remember we now also know the acceleration of the object.

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

Step 5: Substitute the values and calculate the final velocity

\begin{align*} {\vec{v}}_{f} & = {\vec{v}}_{i} + at \\ {\vec{v}}_{f} & = \text{0}\text{ m·s$^{-1}$} + \left(\text{8}\text{ m·s$^{-2}$}\right)\left(\text{4}\text{ s}\right) \\ & = \text{32}\text{ m·s$^{-1}$} \text{ East} \end{align*}

Step 6: Time at half the distance: Find an equation to calculate the time

We can use Equation 3:

\begin{align*} \Delta \vec{x} & = {\vec{v}}_{i} + \frac{1}{2}\vec{a}{t}^{2} \\ \text{32}\text{ m} & = \left(\text{0}\text{ m·s$^{-1}$}\right)t + \frac{1}{2}\left(\text{8}\text{ m·s$^{-2}$}\right){\left(t\right)}^{2} \\ \text{32}\text{ m} & = 0 + \left(\text{4}\text{ m·s$^{-2}$}\right){t}^{2} \\ \text{8}\text{ s$^{2}$} & = {t}^{2} \\ t & = \text{2,83}\text{ s} \end{align*}

Step 7: Distance at half the time: Find an equation to relate the distance and time

Half the time is \(\text{2}\) \(\text{s}\), thus we have \({\vec{v}}_{i}\), \(\vec{a}\) and \(t\) all in the correct units. We can use Equation 3 to get the distance:

\begin{align*} \Delta \vec{x} & = {\vec{v}}_{i}t+\frac{1}{2}a{t}^{2} \\ & = \left(\text{0}\text{ m·s$^{-1}$}\right)\left(\text{2}\text{ s}\right) + \frac{1}{2}\left(\text{8}\text{ m·s$^{-2}$}\right){\left(\text{2}\text{ s}\right)}^{2} \\ & = \text{16}\text{ m} \text{ East} \end{align*}