Answer
a)
(i) Vavg=−32 ft/s
(ii) Vavg=−25.6 ft/s
(iii) Vavg=−24.8 ft/s
(iv) Vavg=−24.16 ft/s
b)
−24 ft/s
Work Step by Step
$t_1=2s$
$v_{avg}=Δy/Δt=\frac{y_2−y_1}{t_2−t_1}=\frac{(40t_2−16(t_2)^2)−(40t_1−16(t_1)^2)}{t_2−t_1}$
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$$
