## Calculus: Early Transcendentals 8th Edition

(a) (i) $3~m/s$ (ii) $2.75~m/s$ (iii) $2.625~m/s$ (iv) $2.525~m/s$ (b) $2.5~m/s$
(a )$s = 1+2t+\frac{1}{4}t^2$ We can find the average velocity over these time intervals: (i) $[1, 3]$ $s(1) = 1+2(1)+\frac{1}{4}(1)^2 = \frac{13}{4}$ $s(3) = 1+2(3)+\frac{1}{4}(3)^2 = \frac{37}{4}$ $v_{ave} = \frac{\frac{37}{4}-\frac{13}{4}}{3-1} = 3~m/s$ (ii) $[1, 2]$ $s(1) = 1+2(1)+\frac{1}{4}(1)^2 = \frac{13}{4}$ $s(2) = 1+2(2)+\frac{1}{4}(2)^2 = \frac{24}{4}$ $v_{ave} = \frac{\frac{24}{4}-\frac{13}{4}}{2-1} = \frac{11}{4} = 2.75~m/s$ (iii) $[1, 1.5]$ $s(1) = 1+2(1)+\frac{1}{4}(1)^2 = \frac{13}{4}$ $s(1.5) = 1+2(1.5)+\frac{1}{4}(1.5)^2 = \frac{73}{16}$ $v_{ave} = \frac{\frac{73}{16}-\frac{52}{16}}{1.5-1} = \frac{21}{8} = 2.625~m/s$ (iv) $[1, 1.1]$ $s(1) = 1+2(1)+\frac{1}{4}(1)^2 = \frac{13}{4}$ $s(1.1) = 1+2(1.1)+\frac{1}{4}(1.1)^2 = 3.5025$ $v_{ave} = \frac{3.5025-3.25}{1.1-1} = 2.525~m/s$ (b) We can find an expression for the instantaneous velocity: $v(t) = \lim\limits_{h \to 0}\frac{s(t+h)-s(t)}{h}$ $=\lim\limits_{h \to 0}\frac{[1+2(t+h)+\frac{1}{4}(t+h)^2]-[1+2t+\frac{1}{4}t^2]}{h}$ $=\lim\limits_{h \to 0}\frac{[1+2t+2h+\frac{1}{4}(t^2+2th+h^2)]-[1+2t+\frac{1}{4}t^2]}{h}$ $=\lim\limits_{h \to 0}\frac{2h+\frac{1}{4}(2th+h^2)}{h}$ $=\lim\limits_{h \to 0}[2+\frac{1}{4}(2t+h)]$ $= 2+\frac{1}{2}t$ When $t = 1$: $v(1) = 2+\frac{1}{2}(1) = 2.5~m/s$