A ship leaves a port at 1:00 PM and sails in the direction N 24° W at the rate of 10 km/p for 30 minutes. It then sails N 66°E for 1.2 kilometers at the same speed. Find its distance and bearing from the port at 2:42 PM.
If the port P is at (0,0)
It turns at Q and ends up at R, then
PQ = 10 km/hr * 1/2 hr= 5km
QR = 1.2 km at 10 km/hr = 0.12 hr = 7:20
total time is 0:30 + 7:20 = 37:20
You don't say what it does between 1:37:20 and 2:42 PM
In any case, use the law of cosines to find distances, and then find bearings as usual.
To find the distance and bearing of the ship from the port at 2:42 PM, we can follow these steps:
Step 1: Calculate the distance traveled in the first leg of the journey.
The ship sails in the direction N 24° W at a speed of 10 km/h for 30 minutes. To calculate the distance traveled, we can use the formula: Distance = Speed × Time.
Time = 30 minutes = 30/60 = 0.5 hours.
Distance1 = 10 km/h × 0.5 h = 5 km.
Step 2: Calculate the distance traveled in the second leg of the journey.
The ship sails in the direction N 66° E for 1.2 kilometers at the same speed. The angle difference between the initial direction (N 24° W) and the new direction (N 66° E) is 90° (66° + 24° - 180°). Since we have a right-angled triangle, we can use the Pythagorean theorem to calculate the distance traveled in this leg.
Distance2 = √(5^2 + 1.2^2) = √(25 + 1.44) = √26.44 ≈ 5.14 km.
Step 3: Calculate the total distance from the port.
The total distance from the port is the sum of Distance1 and Distance2.
Total Distance = 5 km + 5.14 km ≈ 10.14 km.
Step 4: Calculate the bearing from the port.
To calculate the bearing, we use the tangent formula:
Tangent of bearing = Opposite Side/Adjacent Side.
In our case, the opposite side is 1.2 km and the adjacent side is 5 km.
Tangent of bearing = 1.2/5 ≈ 0.24.
Taking the inverse tangent (arctan) of 0.24 will give us the bearing.
Bearing ≈ arctan(0.24) ≈ 13.14°.
Therefore, at 2:42 PM, the ship is approximately 10.14 km away from the port, and its bearing from the port is approximately N 13.14° E.
To find the distance and bearing of the ship from the port at 2:42 PM, we need to break down the ship's journey into two segments and calculate the displacement for each segment.
1. Segment 1: Sailing N 24° W for 30 minutes at 10 km/h.
We first need to convert the sailing time to hours. 30 minutes is equal to 0.5 hours.
Using the formula d = r * t (distance = rate * time), we can calculate the distance covered in this segment:
Distance = 10 km/h * 0.5 hours = 5 km
2. Segment 2: Sailing N 66° E for 1.2 kilometers at the same speed.
The distance is given as 1.2 kilometers.
Now, let's calculate the total displacement using vector addition.
To do this, we need to resolve the two segments into their x and y components.
Segment 1:
The angle is N 24° W. To resolve this angle, we need to break it down into its x and y components.
y_component = 5 km * sin(24°) = 2.04 km (northward)
x_component = 5 km * cos(24°) = 4.55 km (westward)
Segment 2:
The angle is N 66° E.
y_component = 1.2 km * sin(66°) = 1.09 km (northward)
x_component = 1.2 km * cos(66°) = 0.48 km (eastward)
Now, we can add up the x and y components:
Total y_component = 2.04 km + 1.09 km = 3.13 km (northward)
Total x_component = 4.55 km + 0.48 km = 5.03 km (westward)
To find the total displacement, we use the Pythagorean theorem:
Total displacement = √((Total x_component)^2 + (Total y_component)^2)
Total displacement = √((5.03 km)^2 + (3.13 km)^2)
Total displacement = √(25.30 km^2 + 9.78 km^2)
Total displacement = √(35.08 km^2)
Total displacement = 5.92 km (rounded to two decimal places)
To find the bearing of the ship from the port, we can use the inverse tangent function:
Bearing = arctan(Total y_component / Total x_component)
Bearing = arctan(3.13 km / 5.03 km)
Bearing = arctan(0.62)
Bearing = 31.14° (rounded to two decimal places)
Therefore, at 2:42 PM, the ship is approximately 5.92 kilometers away from the port, bearing N 31.14° W.