We can calculate acceleration on a velocity vs. time graph in the same way that we calculate velocity on a position vs. time graph. Acceleration is the rate of change of the velocity vector, , which expresses itself as the slope of the velocity vs. time graph. For a velocity vs. time graph, the acceleration at time t is equal to the slope of the line at t.

What is the acceleration of our ant at t = 2.5 and t = 4? Looking quickly at the graph, we see that the slope of the line at t = 2.5 is zero and hence the acceleration is likewise zero. The slope of the graph between t = 3 and t = 5 is constant, so we can calculate the acceleration at t = 4 by calculating the average acceleration between t = 3 and t = 5:

    The minus sign tells us that acceleration is in the leftward direction, since we’ve defined the y-coordinates in such a way that right is positive and left is negative. At t = 3, the ant is moving to the right at 2 cm/s, so a leftward acceleration means that the ant begins to slow down. Looking at the graph, we can see that the ant comes to a stop at t = 4, and then begins accelerating to the right.

 

 

    Velocity vs. time graphs can also tell us about an object’s displacement. Because velocity is a measure of displacement over time, we can infer that:

 

    Graphically, this means that the displacement in a given time interval is equal to the area under the graph during that same time interval. If the graph is above the t-axis, then the positive displacement is the area between the graph and the t-axis. If the graph is below the t-axis, then the displacement is negative, and is the area between the graph and the t-axis. Let’s look at two examples to make this rule clearer.

 

    First, what is the ant’s displacement between t = 2 and t = 3? Because the velocity is constant during this time interval, the area between the graph and the t-axis is a rectangle of width 1 and height 2.

 

 

    The displacement between t = 2 and t = 3 is the area of this rectangle, which is 1 cm/ss = 2 cm to the right.

Next, consider the ant’s displacement between t = 3 and t = 5. This portion of the graph gives us two triangles, one above the t-axis and one below the t-axis.

 

 

   Both triangles have an area of ¹ /₂(1 s)(2 cm/s) = 1 cm. However, the first triangle is above the t-axis, meaning that displacement is positive, and hence to the right, while the second triangle is below the t-axis, meaning that displacement is negative, and hence to the left. The total displacement between t = 3 and t = 5 is:

 

    In other words, at t = 5, the ant is in the same place as it was at t = 3.