In the given problem, the task involves determining the linear relationship between depth and pressure underwater.

Let's break it down:

Summary:

• The pressure $P$ increases linearly as a diver moves deeper underwater.
• At a depth of 5 meters, the pressure is 1519 hPa.
• At a depth of 20 meters, the pressure is 3036 hPa.
• Pressure $P$ is measured in hPa (hectopascals), and depth $d$ in meters.

Part a) Find a function for $P(d)$:

Since $P(d)$ is a linear function, we can assume it has the form: $P(d) = a \cdot d + b$ Where:

• $a$ is the rate of change of pressure with depth (slope),
• $b$ is the pressure at depth 0 meters (y-intercept).

We are given two points:

• $P(5) = 1519$
• $P(20) = 3036$

We can calculate the slope $a$ using the formula for the slope between two points: $a = \frac{P(20) - P(5)}{20 - 5} = \frac{3036 - 1519}{15} = \frac{1517}{15} = 101.13 \, \text{hPa/m}$

Now, we can substitute $a = 101.13$ into the equation for $P(d)$: $P(d) = 101.13 \cdot d + b$

To find $b$, substitute $d = 5$ and $P(5) = 1519$ into the equation: $1519 = 101.13 \cdot 5 + b$ $1519 = 505.65 + b$ $b = 1519 - 505.65 = 1013.35 \, \text{hPa}$

So, the function for pressure as a function of depth is: $P(d) = 101.13 \cdot d + 1013.35$

Part b) Find the depth where the diver experiences a pressure of 4000 hPa:

We have the equation: $P(d) = 101.13 \cdot d + 1013.35$ Set $P(d) = 4000$ and solve for $d$: $4000 = 101.13 \cdot d + 1013.35$ $4000 - 1013.35 = 101.13 \cdot d$ $2986.65 = 101.13 \cdot d$ $d = \frac{2986.65}{101.13} = 29.53 \, \text{meters}$

Thus, the diver is at a depth of approximately 29.53 meters.

Would you like further details or explanations on any part of this problem?

Here are some related questions to consider:

1. How would the equation change if the initial pressure (at 0 meters) were different?
2. How does the slope represent the physical situation in this problem?
3. Can you express the pressure in another unit, like Pascal (Pa) instead of hectoPascal?
4. How does the pressure change if the diver goes to a depth of 10 meters?
5. What happens to the pressure if the depth doubles?

Tip: Remember that a linear relationship means the pressure increases at a constant rate as the depth increases!