Answer
(a) See the graph
(b) $v_{av}=-1.63~m/s$
(c) $v_{av}=-1.64~m/s$
(d) Neither -1.62 m/s nor -1.66 m/s. We would expect the intantaneous velocity to be closer to -1.64 m/s.
Work Step by Step
(a) See graph.
(b) For $t=0.150~s,$
$x=(-2.0~m/s)\times0.150~s+(3.0~m/s)\times(0.150~s)^{3}=-0.289875~m$.
For $t=0.250~s,$
$x=(-2.0~m/s)\times0.250~s+(3.0~m/s)\times(0.250~s)^{3}=-0.453125~m$.
$v_{av}=\frac{\Delta x}{\Delta t}=\frac{-0.453125~m-(-0.289875~m)}{0.250~s-0.150~s}=-1.63~m/s$.
(c) For $t=0.190~s,$
$x=(-2.0~m/s)\times0.190~s+(3.0~m/s)\times(0.190~s)^{3}=-0.359423~m$.
For $t=0.210~s,$
$x=(-2.0~m/s)\times0.210~s+(3.0~m/s)\times(0.210~s)^{3}=-0.392217~m$.
$v_{av}=\frac{\Delta x}{\Delta t}=\frac{-0.392217~m-(-0.359423~m)}{0.210~s-0.190~s}=-1.64~m/s$.
Notice that we used six significant figures for x (-0.289875 m, -0.453125 m, -0.359423 m and 0.392217 m). It was valid because it was an intermediate step, not the final answer. Read "Round-Off Error", last paragraph, page 8.
(d) The average velocity was computed, in both cases, for a time interval centered on $t=0.200~s$. As the interval of time goes to zero the average velocity approaches the instantaneous velocity (See the instantaneous velocity definition, page 24). In (c) the interval time is smaller than in (b). So, we would expect the instantaneous velocity to be closer to -1.64 m/s.
