Question

In the attached position versus time graph what is the magnitude of average velocity for the entire 19 seconds of the motion. Answer in m/s and the number only to the

nearest 100th (ie. The second position after the decimal point.). Remind yourself of the definition of velocity.

Answer 1

According to the plot, the positions at time t = 0 s and t = 19 s are -1 m and -2 m, respectively. So the average velocity for the 19-s interval is

`v_(\rm ave) = (-2\,\mathrm m - (-1\,\mathrm m))/(19\,\mathrm s) = -\frac1{19}(\rm m)/(\rm s)\approx \boxed{-0.05(\rm m)/(\rm s)}`