Answer
(a) See the graph.
(b) $v_{av}=0.55~m/s$
(c) $v_{av}=0.56~m/s$
(d) $0.56~m/s$
Work Step by Step
(a) See graph.
(b) For $t=0.35~s,$
$x=(2.0~m/s)\times0.35~s+(-3.0~m/s)\times(0.35~s)^{3}=0.5714~m$.
For $t=0.45~s,$
$x=(2.0~m/s)\times0.45~s+(-3.0~m/s)\times(0.45~s)^{3}=0.6266~m$.
$v_{av}=\frac{\Delta x}{\Delta t}=\frac{0.6266~m-0.5714~m}{0.45~s-0.35~s}=0.55~m/s$.
(c) For $t=0.39~s,$
$x=(2.0~m/s)\times0.39~s+(-3.0~m/s)\times(0.39~s)^{3}=0.6020~m$.
For $t=0.41~s$
$x=(2.0~m/s)\times0.41~s+(-3.0~m/s)\times(0.41~s)^{3}=0.6132~m$
$v_{av}=\frac{\Delta x}{\Delta t}=\frac{0.6132~m-0.6020~m}{0.41~s-0.39~s}=0.56~m/s$
Notice that we used four significant figures for x (0.5714 m, 0.6266 m, 0.6020 m and 0.6132 m). It was valid because it was an intermediate step, not the final answer. If, for example, we had used two significant figures in item (b), then
for $t=0.35~s,$
$x=(2.0~m/s)\times0.35~s+(-3.0~m/s)\times(0.35~s)^{3}=0.57~m$;
for $t=0.45~s,$
$x=(2.0~m/s)\times0.45~s+(-3.0~m/s)\times(0.45~s)^{3}=0.63~m;$ and
$v_{av}=\frac{\Delta x}{\Delta t}=\frac{0.63~m-0.57~m}{0.45~s-0.35~s}=0.60~m/s$. Read "Round-Off Error", last paragraph, page 8.
(d) The average velocity was computed, in both cases, for a time interval centered on $t=0.40~s$. 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).
