Answer
a)
$f(1) = 38$ feet
$f(2) = 38$ feet
$f(1) = f(2) = 38$ feet
b)
According to Rolle's Theorem, the velocity is equal to $0$ feet per second when time $t = \frac{3}{2}$ seconds.
Work Step by Step
For part a), plug in $t = 1$ and $t = 2$ into $f(t)$:
$f(1) = -16(1)^{2} + 48(1) + 6 = 38$ feet
$f(2) = -16(2)^{2} + 48(2) + 6 = 38$ feet
Therefore, $f(1) = f(2) = 38$ feet.
For part b), according to Rolle's Theorem, the velocity must be equal to $0$ some time in the interval $(1,2)$. To find the velocity function, you need to take the derivative of the position function, $f(t)$:
$v(t) = f'(t) = -32t + 48 = 0$
$-32t = -48$
$32t = 48$
$t = \frac{48}{32}$
$t = \frac{3}{2}$
Therefore, according to Rolle's Theorem, the velocity is equal to $0$ feet per second when time $t = \frac{3}{2}$ seconds.
