More Motion Graph Examples

The position vs. time graph below describes the motion of an athlete.

1. What is the velocity of the athlete during the first \(\text{4}\) seconds?

2. What is the velocity of the athlete from \(t = \text{4}\text{ s}\) to \(t = \text{7}\text{ s}\)?

Step 1: The velocity during the first \(\text{4}\) seconds

The velocity is given by the gradient of a position vs. time graph. During the first \(\text{4}\) seconds, this is

\begin{align*} \vec{v} & = \frac{\Delta \vec{x}}{\Delta t} \\ & = \frac{\text{4}\text{ m} - \text{0}\text{ m}}{\text{4}\text{ s} - \text{0}\text{ s}} \\ & = \text{1}\text{ m·s$^{-1}$} \end{align*}

Step 2: The velocity during the last 3 seconds

For the last 3 seconds we can see that the displacement stays constant. The graph shows a horizontal line and therefore the gradient is zero. Thus \(v = \text{0}\text{ m·s$^{-1}$}\).

The acceleration vs. time graph for a car starting from rest, is given below. Calculate the velocity of the car and hence draw the velocity vs. time graph.

Step 1: Calculate the velocity values by using the area under each part of the graph.

The motion of the car can be divided into three time sections: \(\text{0}\) – \(\text{2}\) seconds; \(\text{2}\) – \(\text{4}\) seconds and \(\text{4}\) – \(\text{6}\) seconds. To be able to draw the velocity vs. time graph, the velocity for each time section needs to be calculated. The velocity is equal to the area of the square under the graph:

For \(\text{0}\) – \(\text{2}\) seconds:

\begin{align*} {\text{Area}}_{\square } & = l \times b \\ & = \text{2}\text{ s} \times \text{2}\text{ m·s$^{-2}$} \\ & = \text{4}\text{ m·s$^{-1}$} \end{align*}

The velocity of the car is \(\text{4}\) \(\text{m·s$^{-1}$}\) at \(t = \text{2}\text{ s}\).

For \(\text{2}\) – \(\text{4}\) seconds:

\begin{align*} {\text{Area}}_{\square } & = l \times b \\ & = \text{2}\text{ s} \times \text{0}\text{ m·s$^{-2}$} \\ & = \text{0}\text{ m·s$^{-1}$} \end{align*}

The velocity of the car is \(\text{0}\) \(\text{m·s$^{-1}$}\) from \(t = \text{2}\text{ s}\) to \(t = \text{4}\text{ s}\).

For \(\text{4}\) – \(\text{6}\) seconds:

\begin{align*} {\text{Area}}_{\square } & = l \times b \\ & = \text{2}\text{ s} \times -\text{2}\text{ m·s$^{-2}$} \\ & = -\text{4}\text{ m·s$^{-1}$} \end{align*}

The acceleration had a negative value, which means that the velocity is decreasing. It starts at a velocity of \(\text{4}\) \(\text{m·s$^{-1}$}\) and decreases to \(\text{0}\) \(\text{m·s$^{-1}$}\).

Step 2: Now use the values to draw the velocity vs. time graph.

The velocity vs. time graph looks like this: