# Motion Graphs

## Motion Graphs Revision

**Motion Graphs**

There are **two** types of motion graphs: **displacement-time graphs** and **velocity-time graphs**.

**Displacement-time graphs** display the distance from an origin point over a period of time.

**Velocity-time graphs** display the velocity of a particle over a period of time.

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

**Displacement-Time Graphs**

Below is an example of a **Displacement-Time** graph.

Some key properties about these graphs:

- The
**gradient**represents the**velocity**. A**steeper**line indicates a**greater****velocity**. - When the line is horizontal, the object is not moving, (i.e. its
**velocity**is 0).

**Note:**

Be wary of confusing **displacement** and **distance** – **displacement** is a vector quantity involving distance and a direction, **distance** is purely the distance from the origin, without a specified direction.

**Velocity-Time Graphs**

Below is an example of a **Velocity-Time** graph.

Some key properties about these graphs:

- The
**gradient**represents the**acceleration**. If the graph is**increasing**, the object is**accelerating**. If the graph is**decreasing**, the object is**decelerating**. - The
**area**under the graph represents the**distance**travelled.

**Note:**

Be careful when adding all areas under a VT graph. Remember, if the **velocity** is negative, the object is heading backwards, so its **displacement** will **decrease**, but its **distance** travelled will **increase**.

**Forming SUVAT Equations from a VT Graph**

Here is another example of a VT graph.

We can find details on the graph which represent the SUVAT variables.

For example, we have the initial velocity, u = -1\text{ ms}^{-1}, the end velocity, \textcolor{red}{v} = \textcolor{red}{4}\text{ ms}^{-1} and the time taken, \textcolor{purple}{t} = \textcolor{purple}{5}\text{ seconds}. From that information, we can calculate \textcolor{blue}{a} and \textcolor{limegreen}{s}.

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

we have

\textcolor{red}{4} = -1 + \textcolor{purple}{5}\textcolor{blue}{a}\\

\textcolor{blue}{a} = \textcolor{blue}{1}\text{ ms}^{-2}

Using \textcolor{limegreen}{s} = \dfrac{1}{2}(u + \textcolor{red}{v})\textcolor{purple}{t}, we have

\textcolor{limegreen}{s} = \dfrac{1}{2} \times (\textcolor{red}{4} - 1) \times \textcolor{purple}{5}

\textcolor{limegreen}{s} = \textcolor{limegreen}{7.5}\text{ m}

## Motion Graphs Example Questions

**Question 1: **Alex transitions from a slow walk of 1\text{ ms}^{-1} to a run of 2.22\text{ ms}^{-1} over 1\text{ second}. After 30\text{ seconds}, he slows to a stop in 3\text{ seconds}. What distance does he run in this time?

**[2 marks]**

To find the distance, we can find the area of each section under the graph. It can be split into a trapezium, a rectangle and a triangle, so we have

\left( \dfrac{1}{2} \times (1 + 2.22) \times 1\right) + (30 \times 2.22) + \left( \dfrac{1}{2} \times 3 \times 2.22\right)

= 71.54\text{m}

**Question 2: **A train journey from Harrogate to London takes 4\text{ hours}. At t = 20\text{ minutes}, the train stops at Leeds for 10\text{ minutes}, before stopping at Sheffield, Nottingham and Leicester for \text{two minutes} each.

a) For which portion of the journey does the train move fastest?

b) When is it slowest (when not at a station)?

**[2 marks]**

a) By inspection, we see that the graph is steepest (therefore travelling fastest) between the first and second stop, **Leeds to Sheffield.**

b) The graph is flattest (therefore travelling slowest) between the fourth and fifth stop, **Nottingham to Leicester**.

**Question 3: **A cyclist is travelling at 15\text{ ms}^{-1} and accelerates at a constant rate to 25\text{ms}^{-1}.

Using the graph, calculate her acceleration, and the distance travelled in that time.

**[2 marks]**

The change in velocity is 10\text{ ms}^{-1} over 12\text{ seconds}.

This indicates an acceleration of \dfrac{5}{6}\text{ ms}^{-2} = 0.833\text{ ms}^{-2}.

Using s = \dfrac{1}{2}(u + v)t, we have s = \dfrac{1}{2} \times 40 \times 12 = 240\text{ m}.