Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t=30.0 s is approximately 0.24 m/s.

Solution:

We can obviously see from the graph that it is a straight line or approximately a straight line. In this case, the slope is constant.

To get an approximate slope at t=30 s, we can use the values at t=20 s and t=40 s. When t=20\ \text{s}, x=7\ \text{m} and when t=40\ \text{s}, x=12\ \text{m}.

\begin{align*}

\text{slope} & =\frac{\Delta x}{\Delta t} \\
\text{slope} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{slope} & = \frac{12\ \text{m}-7\ \text{m}}{40\ \text{s}-20\ \text{s}} \\
\text{slope} & =0.25\ \text{m/s}

\end{align*}

Although not equal, the computed slope is almost the same with 0.24 m/s. This is due to the fact that values are uncertainties when using graphs. The difference is not really significant for this case.