The problem states that the power $P$ required to propel a ship varies directly as the cube of the speed $s$. This relationship can be expressed as:

$P = k \cdot s^3$

Where:

• $P$ is the power,
• $s$ is the speed,
• $k$ is a constant of proportionality.

Step 1: Find the value of $k$

From the problem, we know:

• $P = 4200$ hp when $s = 14.0$ mi/h.

Substitute these values into the equation $P = k \cdot s^3$:

$4200 = k \cdot (14.0)^3$

First, calculate $14.0^3$:

$14.0^3 = 2744$

Now solve for $k$:

$k = \frac{4200}{2744} \approx 1.53$

Step 2: Use $k$ to find $P$ at $s = 17.0$ mi/h

We now need to find the power $P$ when $s = 17.0$ mi/h. Substitute $k = 1.53$ and $s = 17.0$ into $P = k \cdot s^3$:

$P = 1.53 \cdot (17.0)^3$

First, calculate $17.0^3$:

$17.0^3 = 4913$

Now multiply:

$P = 1.53 \cdot 4913 \approx 7516.89$

Step 3: Round to the nearest hundred

Rounding $7516.89$ to the nearest hundred:

$P \approx 7500 \, \text{hp}.$

Final Answer:

The power required to propel the ship at $17.0 \, \text{mi/h}$ is approximately 7500 hp.

Do you need more clarification or steps on this? Let me know!

Related Questions:

1. What happens to the power if the speed doubles?
2. How do we verify that the relationship $P = k \cdot s^3$ holds true?
3. If $s = 20 \, \text{mi/h}$, what power would be required?
4. How does varying the speed impact fuel efficiency if power increases?
5. What are real-world applications of this power-speed relationship in maritime engineering?

Tip: Remember that in "direct variation," a quantity increases as its variable counterpart increases, following a constant ratio.