Answer
$$\eqalign{
& \left( a \right){\text{Graph velocity}} \cr
& \left( b \right)\ln \left| {t + 1} \right| - 4,{\text{ for }}t \geqslant 0 \cr
& \left( c \right){\text{Graph position}} \cr} $$
Work Step by Step
$$\eqalign{
& v\left( t \right) = \frac{1}{{t + 1}}\,{\text{on the interval }}\left[ {0,8} \right],{\text{ and initial position }}s\left( 0 \right) = - 4 \cr
& \cr
& \left( a \right){\text{Graph shown below }}\left( {{\text{Use GeoGebra}}} \right) \cr
& *{\text{The motion of the object is in the positive direction when }} \cr
& v\left( t \right) > 0.{\text{ then from the graph we see that}} \cr
& v\left( t \right) > 0{\text{ for all the interval }}\left[ {0,8} \right] \cr
& \cr
& \left( b \right){\text{ Position for }}t \geqslant 0 \cr
& *{\text{Using the antiderivative method}} \cr
& {\text{Recall that the velocity of an object at time }}t{\text{ is }} \cr
& v\left( t \right) = s'\left( t \right) \cr
& s'\left( t \right) = v\left( t \right) \cr
& \frac{{ds}}{{dt}} = \frac{1}{{t + 1}} \cr
& {\text{Finding the antiderivative}} \cr
& s\left( t \right) = \int {\frac{1}{{t + 1}}} dt \cr
& s\left( t \right) = \ln \left| {t + 1} \right| + C,{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Using the initial condition }}s\left( 0 \right) = - 2 \cr
& - 4 = \ln \left| {0 + 1} \right| + C \cr
& C = - 4 \cr
& {\text{Substitute }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& s\left( t \right) = \ln \left| {t + 1} \right| - 4,{\text{ for }}t \geqslant 0 \cr
& *{\text{Using the Fundamental theorem of calculus }} \cr
& {\text{Let }}v\left( x \right) = \frac{1}{{x + 1}},{\text{ determine for }}t \geqslant 0 \cr
& {\text{Use theorem 6}}{\text{.1 }}\left( {{\text{page 401}}} \right) \cr
& \underbrace {s\left( t \right)}_{\scriptstyle {\text{position}} \atop
\scriptstyle {\text{at }}t} = \underbrace {s\left( 0 \right)}_{\scriptstyle {\text{initial}} \atop
\scriptstyle {\text{position}}} + \underbrace {\int_0^t {v\left( x \right)} dt}_{\scriptstyle {\text{displacement}} \atop
\scriptstyle {\text{over }}\left[ {0,t} \right]} \cr
& {\text{Therefore}}{\text{,}} \cr
& s\left( t \right) = - 4 + \int_0^t {\frac{1}{{x + 1}}} dx \cr
& {\text{Integrating}} \cr
& s\left( t \right) = - 4 + \left[ {\ln \left| {x + 1} \right|} \right]_0^t \cr
& s\left( t \right) = - 4 + \left[ {\ln \left| {t + 1} \right|} \right] - \ln \left| {0 + 1} \right| \cr
& s\left( t \right) = \ln \left| {t + 1} \right| - 4,{\text{ for }}t \geqslant 0 \cr
& \cr
& \left( c \right){\text{Graph the position function shown below }}\left( {{\text{Use GeoGebra}}} \right) \cr} $$