Answer
(a)$v_{8.0s}=-12\,\mathrm{m/s}$ and $v_{11.0s}=-4.0\,\mathrm{m/s}$
(b) $\Delta x=-18\,\mathrm{m}$
(c) $d=50\,\mathrm{m}$
Work Step by Step
(a) It is clear from the graph that $v_{8.0s}=-12\,\mathrm{m/s}$ and $v_{11.0s}=-4.0\,\mathrm{m/s}$.
(b) Noting that the area under the velocity-time graph is displacement,
Total area under graph = ar(ABC)+ar(DEF)+ar(CDGH)
$={1\over 2}(5.0-1.0)(8.0-0)+{1\over 2}(9.0-6.0)(4.0-12.0)+{1\over 2}(6.0+5.0)(-4.0)$
$=16-12-22$
$=-18\,\mathrm{m}$
Therefore, displacement = $18\,\mathrm{m}$.
(c) To compute the total distance we again calculate the area under the graph but consider the magnitudes of the velocities alone (skipping the signs). And we have,
Total distance $= 16+12+22=50\,\mathrm{m}$.
