Motion at Constant Acceleration
The final situation we will be studying is motion at constant acceleration. We know that acceleration is the rate of change of velocity. So, if we have a constant acceleration, this means that the velocity changes at a constant rate. Let's look at our first example of ...
Motion at Constant Acceleration
The final situation we will be studying is motion at constant acceleration. We know that acceleration is the rate of change of velocity. So, if we have a constant acceleration, this means that the velocity changes at a constant rate.
Let's look at our first example of Vivian waiting at the taxi stop again. A taxi arrived and Vivian got in. The taxi stopped at the stop street and then accelerated in the positive direction as follows: After \(\text{1}\) \(\text{s}\) the taxi covered a distance of \(\text{2.5}\) \(\text{m}\), after \(\text{2}\) \(\text{s}\) it covered \(\text{10}\) \(\text{m}\), after \(\text{3}\) \(\text{s}\) it covered \(\text{22.5}\) \(\text{m}\) and after \(\text{4}\) \(\text{s}\) it covered \(\text{40}\) \(\text{m}\). The taxi is covering a larger distance every second. This means that it is accelerating.
To calculate the velocity of the taxi you need to calculate the gradient of the line at each second:
\begin{align*} \vec{v}_{1s} & = \frac{\Delta \vec{x}}{\Delta t} \\ & = \frac{{\vec{x}}_{f} - {\vec{x}}_{i}}{{t}_{f} - {t}_{i}} \\ & = \frac{\text{5}\text{ m} - \text{0}\text{ m}}{\text{1,5}\text{ s} - \text{0,5}\text{ s}} \\ & = \text{5}\text{ m·s$^{-1}$} \end{align*}\begin{align*} \vec{v}_{2s} & = \frac{\Delta \vec{x}}{\Delta t} \\ & = \frac{{\vec{x}}_{f} - {\vec{x}}_{i}}{{t}_{f} - {t}_{i}} \\ & = \frac{\text{15}\text{ m} - \text{5}\text{ m}}{\text{2,5}\text{ s} - \text{1,5}\text{ s}} \\ & = \text{10}\text{ m·s$^{-1}$} \end{align*}\begin{align*} \vec{v}_{1s} & = \frac{\Delta \vec{x}}{\Delta t} \\ & = \frac{{\vec{x}}_{f} - {\vec{x}}_{i}}{{t}_{f} - {t}_{i}} \\ & = \frac{\text{30}\text{ m} - \text{15}\text{ m}}{\text{3,5}\text{ s} - \text{2,5}\text{ s}} \\ & = \text{15}\text{ m·s$^{-1}$} \end{align*}
From these velocities, we can draw the velocity-time graph which forms a straight line.
The acceleration is the gradient of the \(v\) vs. \(t\) graph and can be calculated as follows:
\begin{align*} a & = \frac{\Delta \vec{v}}{\Delta t} \\ & = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{{t}_{f} - {t}_{i}} \\ & = \frac{\text{15}\text{ m·s$^{-1}$} - \text{5}\text{ m·s$^{-1}$}}{\text{3}\text{ s} - \text{1}\text{ s}} \\ & = \text{5}\text{ m·s$^{-2}$} \end{align*}
The acceleration does not change during the motion (the gradient stays constant). This is motion at constant or uniform acceleration.
The graphs for this situation are shown below:
Graphs for motion with a constant acceleration starting from rest.
Velocity From Acceleration vs. Time Graphs
Just as we used velocity vs. time graphs to find displacement, we can use acceleration vs. time graphs to find the velocity of an object at a given moment in time. We simply calculate the area under the acceleration vs. time graph, at a given time. In the graph below, showing an object at a constant positive acceleration, the increase in velocity of the object after \(\text{2}\) seconds corresponds to the shaded portion.
\begin{align*} v = \text{ area of rectangle } & = a\times \Delta t \\ & = \text{5}\text{ m·s$^{-2}$} \times \text{2}\text{ s} \\ & = \text{10}\text{ m·s$^{-1}$} \end{align*}The velocity of the object at \(t = \text{2}\text{ s}\) is therefore \(\text{10}\) \(\text{m·s$^{-1}$}\).
This lesson is part of:
One-Dimensional Motion