## Calculus 8th Edition

$t_1=2s$ $v_{avg}=Δy/Δt=\frac{y_2−y_1}{t_2−t_1}=\frac{(40t_2−16(t_2)^2)−(40t_1−16(t_1)^2)}{t_2−t_1}$ a) (i) $t_1=2s$ $t_2=2.5s$ $v_{avg}=−32$ ft/s (ii) $t_1=2s$ $t_2=2.1s$ $v_{avg}=−25.6$ ft/s (iii) $t_1=2s$ $t_2=2.05s$ $v_{avg}=−24.8$ ft/s (iv) $t_1=2s$ $t_2=2.01s$ $v_{avg}=−24.16$ ft/s b) To find the instantaneous velocity at a specific time, we take the first derivative of the position equation. $y=40t−16t_2$ $y′=40−32t$ $y′(2)=40−32(2)$ $y′(2)=−24 ft/s$\$