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

$v_{\rm ave} = \dfrac{-2\,\mathrm m - (-1\,\mathrm m)}{19\,\mathrm s} = -\dfrac1{19}\dfrac{\rm m}{\rm s}\approx \boxed{-0.05\dfrac{\rm m}{\rm s}}$