Question

question 16 - 1 point if a brick is dropped from a height of 81 ft, its position after t seconds is given by the function s(t)= - 16t^2 + 81. find the time, guaranteed by the mean value theorem, when the instantaneous velocity of the brick equals its average velocity over the complete fall. (enter an exact answer.) provide your answer below: time = seconds

Explanation:

Step1: Recall velocity - related formulas

The instantaneous velocity $v(t)$ is the derivative of the position - function $s(t)$. So, if $s(t)=-16t^{2}+81$, then $v(t)=s^\prime(t)=-32t$ using the power rule $\frac{d}{dt}(at^{n}) = nat^{n - 1}$. The average velocity over the interval $[0,t]$ is $\frac{s(t)-s(0)}{t - 0}$. Since $s(0)=-16(0)^{2}+81 = 81$ and $s(t)=-16t^{2}+81$, the average velocity $\overline{v}=\frac{(-16t^{2}+81)-81}{t}=\frac{-16t^{2}}{t}=-16t$.

Step2: Set instantaneous velocity equal to average velocity

We want to find when $v(t)=\overline{v}$. So, set $-32t=-16t$. This equation is not correct. We should use the Mean - Value Theorem which states that if $y = s(t)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $c\in(a,b)$ such that $s^\prime(c)=\frac{s(b)-s(a)}{b - a}$. Here, $a = 0$ and we want to find the time $t$ when the condition is met. The position function is $s(t)=-16t^{2}+81$. The brick hits the ground when $s(t)=0$. So, $-16t^{2}+81 = 0$.

Step3: Solve for the time when the brick hits the ground

Rearrange the equation $-16t^{2}+81 = 0$ to get $16t^{2}=81$. Then $t^{2}=\frac{81}{16}$. Taking the square root of both sides, we have $t=\frac{9}{4}$ (we take the positive value since time $t\geq0$ in this context).

Answer:

$\frac{9}{4}$