Answer
a) $s=-16t^2+80$
b) See graph
c) $-48$
d) Negative slope
e) $y=-48t+112$
f) See graph
Work Step by Step
a) The function representing the position is:
$s=-16t^2+v_0t+s_0$
where $v_0$ is the velocity and $s_0$ is the height from which the object is thrown.
We are given:
$s_0=80$
$v_0=0$
The equation of the position is:
$s=-16t^2+80t$
b) Graph the function.
c) Determine the average rate of change from $t_1=1$ to $t_2=2$:
$\dfrac{s(t_2)-s(t_1)}{t_2-t_1}=\dfrac{s(2)-s(1)}{2-1}$
$=\dfrac{[-16(2^2)+80]-[-16(1^2)+80]}{1}$
$=\dfrac{-48}{1}$
$=-48$
d) Notice that the slope of the secant line is negative.
e) Determine the equation of the secant using the slope $m=-48$ and the point $(1,s(1))=(1,-16(1^2)+80)=(1,64) $:
$y-64=-48(t-1)$
$y=-48t+48+64$
$y=-48t+112$
f) Graph the secant line:
