Answer is: option1

1 knot

Solution:

\( D^2 = W^2 + S^2 \)

where \( D \) is the distance between the two ships, \( W(t) \) is the westward distance, and \( S(t) \) is the southward distance.

Differentiate both sides with respect to \( t \):

\( \frac{d}{dt} (D^2) = \frac{d}{dt} (W^2 + S^2) \)

Using the chain rule:

\( 2D \frac{dD}{dt} = 2W \frac{dW}{dt} + 2S \frac{dS}{dt} \)

Cancel the factor of 2:

\( D \frac{dD}{dt} = W \frac{dW}{dt} + S \frac{dS}{dt} \)

Solving for \( \frac{dD}{dt} \):

\( \frac{dD}{dt} = \frac{W \frac{dW}{dt} + S \frac{dS}{dt}}{D} \)

We are given:

\( W(1) = 5, W'(1) = 0.5 \)

\( S(1) = 4, S'(1) = 1 \)

Compute \( D(1) \):

\( D(1) = \sqrt{W(1)^2 + S(1)^2} = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \)

Now substitute:

\( \frac{dD}{dt} = \frac{(5 \times 0.5) + (4 \times 1)}{\sqrt{41}} \)

\( \frac{dD}{dt} = \frac{2.5 + 4}{\sqrt{41}} \)

\( \frac{dD}{dt} = \frac{6.5}{\sqrt{41}} \)

Approximating \( \sqrt{41} \approx 6.4 \):

\( \frac{dD}{dt} \approx \frac{6.5}{6.4} \approx 1.02 \)

Final Answer is : 1 knot