Question

A fishing boat leaves a marina and follows a course of s61w at 8 knots for 15min . then the boat changes to a new course of s38w at 3knots for 2hr. round intermediate steps to four decimal places.

Answer 1

The distance between the fishing boat and Marina is approximately 7.2698 knots.

To find the total distance traveled by the fishing boat, we can break down the motion into its horizontal and vertical components. The boat is initially moving in a direction S61°W at 6 knots for 15 minutes, and then it changes to a new course of S38°W at 3 knots for 2 hours.

Let's perform the calculations step by step:

• Step 1: Calculate the displacement during the first leg of the journey.

`\[ d_1 = 6 \, \text{knots} \cdot (15)/(60) \, \text{hours} \]`

= 1.5 knots

• Step 2: Calculate the horizontal and vertical components of the displacement during the first leg.

The angle is S61°W.

`\[ d_(1h) = 1.5 \, \text{knots} \cdot \cos(61^o) \]` = 0.7071

`\[ d_(1v) = 1.5 \, \text{knots} \cdot \sin(61^o) \]` = 1.2764

• Step 3: Calculate the displacement during the second leg of the journey.

`\[ d_2 = 3 \, \text{knots} \cdot 2 \, \text{hours} \]`

`\[ d_2 = 6 \, \text{knots} \]`

• Step 4: Calculate the horizontal and vertical components of the displacement during the second leg.

The angle is S38°W.

`\[ d_(2h) = 6 \, \text{knots} \cdot \cos(38^o) \]` = 4.6107

`\[ d_(2v) = 6 \, \text{knots} \cdot \sin(38^o) \]` = 3.6621

• Step 5: Sum up the horizontal and vertical components of the displacements.

`\[ \text{Total horizontal displacement} = d_(1h) + d_(2h) \]`

`\[ \text{Total vertical displacement} = d_(1v) + d_(2v) \]`

`\[ \text{Total horizontal displacement} \approx 5.3178 \, \text{knots} \]`

`\[ \text{Total vertical displacement} \approx 4.9385 \, \text{knots} \]`

• Step 6: Use the Pythagorean theorem to find the total distance.

`\[ \text{Total distance} = \sqrt{(\text{Total horizontal displacement})^2 + (\text{Total vertical displacement})^2} \]`

`\[ \text{Total distance} \approx √((5.3178)^2 + (4.9385)^2) \]`

`\[ \text{Total distance} \approx √(28.3419 + 24.3874) \]`

`\[ \text{Total distance} \approx √(52.7293) \]`

`\[ \text{Total distance} \approx 7.2698 \, \text{knots} \]`

Therefore, the distance between the fishing boat and Marina is approximately 7.2698 knots.

The complete question:

A fishing boat leaves a marina and follows a course of S61°W at 8 knots for 15 min. Then the boat changes to a new course of S38°W at 3 knots for 2 hrs. How far is the boat from the marina?