Answer
a) $s=-16t^2+64t+6$
b) See graph
c) $16$
d) Positive slope
e) $y=16t+6$
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 where the object is thrown.
We are given:
$s_0=6$
$v_0=64$
The equation of the position is:
$s=-16t^2+64t+6$
b) Graph the function.
c) Determine the average rate of change from $t_1=0$ to $t_3$:
$\dfrac{s(t_2)-s(t_1)}{t_2-t_1}=\dfrac{s(3)-s(0)}{3-0}$
$=\dfrac{[-16(3^2)+64(3)+6]-[-16(0^2)+64(0)+6]}{3}$
$=\dfrac{48}{3}$
$=16$
d) Notice that the slope of the secant line is positive.
e) Determine the equation of the secant using the slope $m=16$ and the point $(0,s(0))=(0,-16(0^2)+64(0)+6)=(0,6) $:
$y-6=16(t-0)$
$y=16t+6$
f) Graph the secant line:
