Question

consider the following figure. a diver stands at the very end of a diving board before beginning a dive. the diving board is l feet long. the board is deflected d at a position s feet from the stationary end. the deflection d of the board at a position s feet from the stationary end is given by d = cs²(3l - s) for 0 ≤ s ≤ l, where l is the length of the board and c is a positive constant that depends on the weight of the diver and on the physical properties of the board. suppose the board is 10 feet long. (a) if the deflection at the end of the board is 1 foot, find c. (b) show that the deflection is 1/2 foot somewhere between s = 6.8 and s = 6.9. (round your answers to four decimal places.) d(6.8)= d(6.9)=

Explanation:

Step1: Identify given values

Given $L = 10$ feet. At the end of the board $s = L=10$ feet. The deflection formula is $d = cs^{2}(3L - s)$ for $0\leq s\leq L$.

Step2: Substitute values into formula

When $s = L = 10$, we have $d = c\times10^{2}(3\times10 - 10)$. Simplify the expression inside the parentheses: $3\times10 - 10=20$. So $d = c\times100\times20=2000c$. And we know that when $s = 10$, $d = 1$ foot. Then $1 = 2000c$. Solve for $c$: $c=\frac{1}{2000}=0.0005$.

Step3: Calculate $d(6.8)$

Now that $c = 0.0005$, use the formula $d(s)=cs^{2}(3L - s)$ with $s = 6.8$ and $L = 10$. Substitute the values: $d(6.8)=0.0005\times6.8^{2}(3\times10 - 6.8)$. First, calculate $6.8^{2}=46.24$ and $3\times10 - 6.8 = 23.2$. Then $d(6.8)=0.0005\times46.24\times23.2$. $0.0005\times46.24 = 0.02312$, and $0.02312\times23.2 = 0.536384\approx0.5364$.

Step4: Calculate $d(6.9)$

Use the formula $d(s)=cs^{2}(3L - s)$ with $s = 6.9$ and $L = 10$. Substitute the values: $d(6.9)=0.0005\times6.9^{2}(3\times10 - 6.9)$. First, calculate $6.9^{2}=47.61$ and $3\times10 - 6.9 = 23.1$. Then $d(6.9)=0.0005\times47.61\times23.1$. $0.0005\times47.61=0.023805$, and $0.023805\times23.1 = 0.5498955\approx0.5499$.

Answer:

$d(6.8)\approx0.5364$
$d(6.9)\approx0.5499$