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QuestionFind the distance and bearing from Corpus Christi to the Coast Guard cutter after it travels at 30 knots for 4 hours on given courses.

Studdy Solution

STEP 1

Assumptions1. The speed of the Coast Guard cutter Angelica is30 knots. . The first course is 95∘95^{\circ} for hours.
3. The second course is 185∘185^{\circ} for hours.
4. The bearing is measured clockwise from north.
5. The distance is calculated in nautical miles, since the speed is given in knots.

STEP 2

First, we need to find the distance travelled in each course. We can do this by multiplying the speed by the time.
Distance=Speed×TimeDistance = Speed \times Time

STEP 3

Now, plug in the given values for the speed and time to calculate the distance for the first course.
Distance1=30 knots×2 hoursDistance1 =30\, knots \times2\, hours

STEP 4

Calculate the distance travelled in the first course.
Distance1=30 knots×2 hours=60 nautical milesDistance1 =30\, knots \times2\, hours =60\, nautical\, miles

STEP 5

Now, plug in the given values for the speed and time to calculate the distance for the second course.
Distance2=30 knots×2 hoursDistance2 =30\, knots \times2\, hours

STEP 6

Calculate the distance travelled in the second course.
Distance2=30 knots×2 hours=60 nautical milesDistance2 =30\, knots \times2\, hours =60\, nautical\, miles

STEP 7

The problem is a vector addition problem. We have two vectors, one for each course, and we need to add them together. The first vector is60 nautical miles at a bearing of 95∘95^{\circ}, and the second vector is60 nautical miles at a bearing of 185∘185^{\circ}. The angle between the two vectors is 185∘−95∘=90∘185^{\circ} -95^{\circ} =90^{\circ}.

STEP 8

To add the vectors, we will use the Pythagorean theorem because the vectors are at a right angle to each other.
Resultant Distance=Distance12+Distance22Resultant\, Distance = \sqrt{{Distance1}^2 + {Distance2}^2}

STEP 9

Plug in the values for the distances to calculate the resultant distance.
Resultant Distance=602+602Resultant\, Distance = \sqrt{{60}^2 + {60}^2}

STEP 10

Calculate the resultant distance.
Resultant Distance=602+602=7200=84.85 nautical milesResultant\, Distance = \sqrt{{60}^2 + {60}^2} = \sqrt{7200} =84.85\, nautical\, miles

STEP 11

To find the bearing, we use the tangent of the angle formed by the two vectors. The tangent of this angle is the ratio of the second distance to the first distance.
Bearing=tan⁡−(DistanceDistance)Bearing = \tan^{-}\left(\frac{Distance}{Distance}\right)

STEP 12

Plug in the values for the distances to calculate the bearing.
Bearing=tan⁡−(6060)Bearing = \tan^{-}\left(\frac{60}{60}\right)

STEP 13

Calculate the bearing.
Bearing=tan⁡−(6060)=45∘Bearing = \tan^{-}\left(\frac{60}{60}\right) =45^{\circ}

STEP 14

The bearing calculated is relative to the first course. To find the absolute bearing, we add the bearing of the first course to the calculated bearing.
Absolute Bearing=First Course Bearing+Calculated BearingAbsolute\, Bearing = First\, Course\, Bearing + Calculated\, Bearing

STEP 15

Plug in the values for the bearings to calculate the absolute bearing.
Absolute Bearing=95∘+45∘Absolute\, Bearing =95^{\circ} +45^{\circ}

STEP 16

Calculate the absolute bearing.
Absolute Bearing=95∘+45∘=140∘Absolute\, Bearing =95^{\circ} +45^{\circ} =140^{\circ}The Coast Guard cutter Angelica is84.85 nautical miles away from the Corpus Christi port at a bearing of 140∘140^{\circ}.

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