Chapter 4 - Integration - 4.7 Rectilinear Motion Revisited Using Integration - Exercises Set 4.7 - Page 329: 7

$$\eqalign{ & \left( {\text{a}} \right)s\left( t \right) = \frac{3}{2}{t^2} + t - 4 \cr & \left( {\text{b}} \right)s\left( t \right) = \frac{1}{t} + t \cr} $$

$$\eqalign{ & \left( {\text{a}} \right){\text{ }}v\left( t \right) = 3t + 1;{\text{ }}s\left( 2 \right) = 4 \cr & s\left( t \right) = \int {v\left( t \right)} dt \cr & s\left( t \right) = \int {\left( {3t + 1} \right)} dt \cr & s\left( t \right) = \frac{3}{2}{t^2} + t + C \cr & {\text{Use the initial condition }}s\left( 2 \right) = 4 \cr & 4 = \frac{3}{2}{\left( 2 \right)^2} + 2 + C \cr & C = - 4 \cr & {\text{Then,}} \cr & s\left( t \right) = \frac{3}{2}{t^2} + t - 4 \cr & \cr & \left( {\text{b}} \right){\text{ }}a\left( t \right) = 2{t^{ - 3}};{\text{ }}v\left( 1 \right) = 0;{\text{ }}s\left( 1 \right) = 2; \cr & v\left( t \right) = \int {a\left( t \right)} dt \cr & v\left( t \right) = \int {2{t^{ - 3}}} dt \cr & v\left( t \right) = - {t^{ - 2}} + C \cr & {\text{Use the initial condition }}v\left( 1 \right) = 0 \cr & 0 = - {\left( 1 \right)^{ - 2}} + C \cr & C = 1 \cr & {\text{Then,}} \cr & v\left( t \right) = - {t^{ - 2}} + 1 \cr & s\left( t \right) = \int {v\left( t \right)} dt \cr & s\left( t \right) = \int {\left( { - {t^{ - 2}} + 1} \right)} dt \cr & s\left( t \right) = \frac{1}{t} + t + C \cr & {\text{Use the initial condition }}s\left( 1 \right) = 2 \cr & 2 = \frac{1}{1} + 1 + C \cr & C = 0 \cr & {\text{Then,}} \cr & s\left( t \right) = \frac{1}{t} + t \cr} $$