Chapter 4 - Integration - 4.7 Rectilinear Motion Revisited Using Integration - Exercises Set 4.7 - Page 329: 8

$$\eqalign{ & \left( {\text{a}} \right)s\left( t \right) = \frac{3}{5}{t^{5/3}} + \frac{{96}}{5} \cr & \left( {\text{b}} \right)s\left( t \right) = \frac{4}{{15}}{t^{5/2}} - \frac{{13}}{3}t + \frac{{19}}{5} \cr} $$

$$\eqalign{ & \left( {\text{a}} \right){\text{ }}v\left( t \right) = {t^{2/3}};{\text{ }}s\left( 8 \right) = 0 \cr & s\left( t \right) = \int {v\left( t \right)} dt \cr & s\left( t \right) = \int {\left( {{t^{2/3}}} \right)} dt \cr & s\left( t \right) = \frac{3}{5}{t^{5/3}} + C \cr & {\text{Use the initial condition }}s\left( 8 \right) = 0 \cr & 0 = \frac{3}{5}{\left( 8 \right)^{5/3}} + C \cr & C = \frac{{96}}{5} \cr & {\text{Then,}} \cr & s\left( t \right) = \frac{3}{5}{t^{5/3}} + \frac{{96}}{5} \cr & \cr & \left( {\text{b}} \right){\text{ }}a\left( t \right) = \sqrt t ;{\text{ }}v\left( 4 \right) = 1;{\text{ }}s\left( 4 \right) = - 5; \cr & v\left( t \right) = \int {a\left( t \right)} dt \cr & v\left( t \right) = \int {{t^{1/2}}} dt \cr & v\left( t \right) = \frac{2}{3}{t^{3/2}} + C \cr & {\text{Use the initial condition }}v\left( 1 \right) = 0 \cr & 1 = \frac{2}{3}{\left( 4 \right)^{3/2}} + C \cr & C = - \frac{{13}}{3} \cr & {\text{Then,}} \cr & v\left( t \right) = \frac{2}{3}{t^{3/2}} - \frac{{13}}{3} \cr & s\left( t \right) = \int {v\left( t \right)} dt \cr & s\left( t \right) = \int {\left( {\frac{2}{3}{t^{3/2}} - \frac{{13}}{3}} \right)} dt \cr & s\left( t \right) = \frac{4}{{15}}{t^{5/2}} - \frac{{13}}{3}t + C \cr & {\text{Use the initial condition }}s\left( 4 \right) = - 5 \cr & - 5 = \frac{4}{{15}}{\left( 4 \right)^{5/2}} - \frac{{13}}{3}\left( 4 \right) + C \cr & - 5 = \frac{{128}}{{15}} - \frac{{52}}{3} + C \cr & C = \frac{{19}}{5} \cr & {\text{Then,}} \cr & s\left( t \right) = \frac{4}{{15}}{t^{5/2}} - \frac{{13}}{3}t + \frac{{19}}{5} \cr} $$