Answer
a)
(i) $v_{avg}=-32\thinspace \frac{ft}{s}$
(ii) $v_{avg}=-25.6\thinspace \frac{ft}{s}$
(iii) $v_{avg}=-24.8\thinspace \frac{ft}{s}$
(iv) $v_{avg}=-24.16\thinspace \frac{ft}{s}$
b)
$-24\thinspace \frac{ft}{s}$
Work Step by Step
$t_1=2\thinspace s$
$v_{avg}=\frac{\Delta y}{\Delta t}=\frac{y_2-y_1}{t_2-t_1}=\frac{(40t_2-16t_2^2)-(40t_1-16t_1^2)}{t_2-t_1}$
a)
(i)
$t_1=2\thinspace s$
$t_2=2.5\thinspace s$
$v_{avg}=-32\thinspace \frac{ft}{s}$
(ii)
$t_1=2\thinspace s$
$t_2=2.1\thinspace s$
$v_{avg}=-25.6\thinspace \frac{ft}{s}$
(iii)
$t_1=2\thinspace s$
$t_2=2.05\thinspace s$
$v_{avg}=-24.8\thinspace \frac{ft}{s}$
(iv)
$t_1=2\thinspace s$
$t_2=2.01\thinspace s$
$v_{avg}=-24.16\thinspace \frac{ft}{s}$
b) To find the instantaneous velocity at a specific time, we take the first derivative of the position equation.
$y=40t-16t^{2}$
$y'=40-32t$
$y'(2)=40-32(2)$
$y'(2)=-24\thinspace \frac{ft}{s}$
