Question

12.3: up and down with the piccards1. challenger deep is the deepest known point in the ocean, at 35,814 feet below sea level. in 1960, jacques piccard and don walsh rode down in the trieste and became the first people to visit the challenger deep.a. if sea level is represented by 0 feet, explain how you can represent the depth of a submarine descending from sea level to the bottom of challenger deep.b. triestes descent was a change in depth of -3 feet per second. we can use the relationship $y = -3x$ to model this, where $y$ is the depth (in feet) and $x$ is the time (in seconds). using this model, how much time would the trieste take to reach the bottom?c. it took the trieste 3 hours to ascend back to sea level. this can be modeled by a different relationship $y = kx$. what is the value of $k$ in this situation?

Explanation:

Step1: (Part a) Define sea level baseline

Sea level is 0 feet. Depths below sea level are represented as negative values. The depth of Challenger Deep is $\boldsymbol{-35814}$ feet (negative because it is below sea level).

Step2: (Part b) Set depth equal to model

We know the final depth $y = -35814$ feet, and the model is $y = -3x$. Substitute $y$ into the equation:
$$-35814 = -3x$$

Step3: (Part b) Solve for time $x$

Divide both sides by $-3$ to isolate $x$:
$$x = \frac{-35814}{-3} = 11938$$

Step4: (Part c) Convert time to seconds

3 hours = $3 \times 60 \times 60 = 10800$ seconds. The final depth (starting point) is $y = -35814$ feet, and the final position is $y=0$. The total change in depth is $0 - (-35814) = 35814$ feet. Use the model $y = kx$, substitute $y=35814$ and $x=10800$:
$$35814 = k \times 10800$$

Step5: (Part c) Solve for $k$

Calculate $k$:
$$k = \frac{35814}{10800} = 3.3161\overline{6}$$

Answer:

a. Since sea level is 0 feet, depths below it are negative. The depth of Challenger Deep is represented as $\boldsymbol{-35814}$ feet.
b. $\boldsymbol{11938}$ seconds
c. $\boldsymbol{k = \frac{35814}{10800} = 3.316}$ (rounded to 3 decimal places, or exact value $\frac{5969}{1800}$)