Velocity-Time Graph Example

The velocity vs. time graph of a truck is plotted below. Calculate the distance and displacement of the truck after \(\text{15}\) seconds.

Step 1: Decide how to tackle the problem

We are asked to calculate the distance and displacement of the car. All we need to remember here is that we can use the area between the velocity vs. time graph and the time axis to determine the distance and displacement.

Step 2: Determine the area under the velocity vs. time graph

Break the motion up: \(\text{0}\) – \(\text{5}\) seconds, \(\text{5}\) – \(\text{12}\) seconds, \(\text{12}\) – \(\text{14}\) seconds and \(\text{14}\) – \(\text{15}\) seconds.

For \(\text{0}\) – \(\text{5}\) seconds: The displacement is equal to the area of the triangle on the left:

\begin{align*} {\text{Area}}_{△} & = \frac{1}{2}b\times h \\ & = \frac{1}{2}\times \text{5}\text{ s} \times \text{4}\text{ m·s$^{-1}$} \\ & = \text{10}\text{ m} \end{align*}

For \(\text{5}\) – \(\text{12}\) seconds: The displacement is equal to the area of the rectangle:

\begin{align*} {\text{Area}}_{\square } & = l \times b \\ & = \text{7}\text{ s} \times \text{4}\text{ m·s$^{-1}$} \\ & = \text{28}\text{ m} \end{align*}

For \(\text{12}\) – \(\text{14}\) seconds the displacement is equal to the area of the triangle above the time axis on the right:

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

For \(\text{14}\) – \(\text{15}\) seconds the displacement is equal to the area of the triangle below the time axis:

\begin{align*} {\text{Area}}_{△} & = \frac{1}{2}b\times h \\ & = \frac{1}{2}\times \text{1}\text{ s} \times \text{2}\text{ m·s$^{-1}$} \\ & = \text{1}\text{ m} \end{align*}

Step 3: Determine the total distance of the car

Now the total distance of the car is the sum of all of these areas:

\begin{align*} D & = \text{10}\text{ m} + \text{28}\text{ m} + \text{4}\text{ m} + \text{1}\text{ m} \\ & = \text{43}\text{ m} \end{align*}

Step 4: Determine the total displacement of the car

Now the total displacement of the car is just the sum of all of these areas. HOWEVER, because in the last second (from \(t = \text{14}\text{ s}\) to \(t = \text{15}\text{ s}\)) the velocity of the car is negative, it means that the car was going in the opposite direction, i.e. back where it came from! So, to find the total displacement, we have to add the first \(\text{3}\) areas (those with positive displacements) and subtract the last one (because it is a displacement in the opposite direction).

\begin{align*} \Delta \vec{x} & = \text{10}\text{ m} + \text{28}\text{ m} + \text{4}\text{ m} - \text{1}\text{ m} \\ & = \text{41}\text{ m} \text{ in the positive direction} \end{align*}