Chapter 6 - Applications of Integration - 6.1 Velocity and Net Change - 6.1 Exercises - Page 407: 12

$$\eqalign{ & a.{\text{ Positive direction for }}\left( {1,4} \right] \cr & {\text{ Negative direction for }}\left( {0,1} \right]{\text{ and }}\left( {4,5} \right] \cr & b.{\text{ }}\frac{5}{6}{\text{m}} \cr & c.{\text{ }}\frac{{49}}{6}{\text{m}} \cr} $$

$$\eqalign{ & {\text{Let }}v\left( t \right) = - {t^2} + 5t - 4{\text{ on the interval }}0 \leqslant t \leqslant 5 \cr & \cr & a.{\text{ The graph of the velocity of the given interval is shown}} \cr & {\text{below }}\left( {{\text{Graph using 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{ on the interval }}\left( {1,4} \right] \cr & *{\text{The motion of the object is in the negative direction when }} \cr & v\left( t \right) < 0.{\text{ then from the graph we see that}} \cr & v\left( t \right) < 0{\text{ on the intervals }}\left( {0,1} \right]{\text{ and }}\left( {4,5} \right] \cr & \cr & b.{\text{ The displacement of the object between }}t = a{\text{ and }}t = b \cr & {\text{where }}a > b{\text{ is :}} \cr & s\left( b \right) - s\left( a \right) = \int_a^b {v\left( t \right)} dt,{\text{ }}\left( {{\text{see page 399 on the book}}} \right) \cr & {\text{Therefore}}{\text{,}} \cr & {\text{For the given interval }}0 \leqslant t \leqslant 5 {\text{ }} \cr & s\left( 5 \right) - s\left( 0 \right) = \int_0^5 {\left( { - {t^2} + 5t - 4} \right)} dt \cr & {\text{Integrating}} \cr & s\left( 5 \right) - s\left( 0 \right) = \left[ { - \frac{1}{3}{t^3} + \frac{5}{2}{t^2} - 4t} \right]_0^5 \cr & s\left( 5 \right) - s\left( 0 \right) = \left[ { - \frac{1}{3}{{\left( 5 \right)}^3} + \frac{5}{2}{{\left( 5 \right)}^2} - 4\left( 5 \right)} \right] - \left[ 0 \right] \cr & s\left( 5 \right) - s\left( 0 \right) = \frac{5}{6} \cr & {\text{The displacement is }}\frac{5}{6}{\text{m}} \cr & \cr & c.{\text{ The distance traveled of the given interval is given by}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \int_0^5 {\left| { - {t^2} + 5t - 4} \right|} dt \cr & {\text{By the definition of absolute value and using the graph}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = - \int_0^1 {\left( { - {t^2} + 5t - 4} \right)} dt + \int_1^4 {\left( { - {t^2} + 5t - 4} \right)} dt \cr & - \int_4^5 {\left( { - {t^2} + 5t - 4} \right)} dt \cr & {\text{Integrating }} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = - \left[ { - \frac{1}{3}{t^3} + \frac{5}{2}{t^2} - 4t} \right]_0^1 + \left[ { - \frac{1}{3}{t^3} + \frac{5}{2}{t^2} - 4t} \right]_2^4 \cr & - \left[ { - \frac{1}{3}{t^3} + \frac{5}{2}{t^2} - 4t} \right]_4^5 \cr & {\text{Evaluating and simplifying}} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \frac{{11}}{6} + \frac{9}{2} + \frac{{11}}{6} \cr & \int_a^b {\left| {v\left( t \right)} \right|} dt = \frac{{49}}{6} \cr} $$