From the x vs. t (position vs. time) graph described in the source text, where the line goes from (t = 0 s, x = 10 m) to (t = 50 s, x = 40 m), what is the average velocity $v_{av}$ of the particle over $\Delta t = 50.0\,\text{s}$ (in m/s)?

What the graph tells you

On an $x$ vs. $t$ (position vs. time) graph, the average velocity over a time interval equals the change in position divided by the change in time. For a straight line segment, this is also the line’s slope.

Read the two endpoints from the described figure

The line goes from:

• Initial point: $t_1=0\,\text{s}$, $x_1=10\,\text{m}$
• Final point: $t_2=50\,\text{s}$, $x_2=40\,\text{m}$

Compute average velocity (slope)

Compute the changes:

$$\Delta x = x_2-x_1 = 40-10 = 30\,\text{m}$$ $$\Delta t = t_2-t_1 = 50-0 = 50\,\text{s}$$

Then:

$$v_{av}=\frac{\Delta x}{\Delta t}=\frac{30\,\text{m}}{50\,\text{s}}=0.600\,\text{m/s}$$

Sign check

Because position increases from $10\,\text{m}$ to $40\,\text{m}$, $\Delta x>0$, so the average velocity is positive.