Answer
a) $s=-16t^2+120t$
b) See graph
c) $-8$
d) Negative slope
e) $y=-8t+240$
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:
$s_0=0$
$v_0=120$
The equation of the position is:
$s=-16t^2+120t$
b) Graph the function.
c) Determine the average rate of change from $t_1=3$ to $t_2=5$:
$\dfrac{s(t_2)-s(t_1)}{t_2-t_1}=\dfrac{s(5)-s(3)}{5-3}$
$=\dfrac{[-16(5^2)+120(5)]-[-16(3^2)+120(3)]}{2}$
$=\dfrac{-16}{2}$
$=-8$
d) Notice that the slope of the secant line is negative.
e) Determine the equation of the secant using the slope $m=-8$ and the point $(3,s(3))=(3,-16(3^2)+120(3))=(3,216) $:
$y-216=-8(t-3)$
$y=-8t+24+216$
$y=-8t+240$
f) Graph the secant line:
