Question

two ships leave a harbour. the first leaves at 12:00pm at a speed of 20km/hr on a course 30 degrees east of north. the second leaves at 12:30pm at a speed of 15km/hr on a course of 20 degrees east of south. at 2:30pm, what is the distance between the ships? use the distance, speed, time triangle in your answer, along with basic eleventh grade trigonometric ratios.

Answer 1

The distance between the two ships at 2:30 pm is 58.31 km.

Step-by-step explanation:

To find the distance between the two ships at 2:30 pm, we first need to find the positions of both ships at that time. Let's start with the first ship:

The first ship leaves at 12:00 pm and travels at a speed of 20 km/hr. By 2:30 pm, it has been sailing for 2.5 hours, so it has traveled a distance of 20 km/hr * 2.5 hrs = 50 km in the direction 30 degrees east of north.

Now let's find the position of the second ship:

The second ship leaves at 12:30 pm and travels at a speed of 15 km/hr. At 2:30 pm, it has been sailing for 2 hours. It has traveled a distance of 15 km/hr * 2 hrs = 30 km in the direction 20 degrees east of south.

Now that we have the positions of both ships, we can use the distance formula to find the distance between them:

Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Using the positions we found, the distance is:

Distance = sqrt((50 km)^2 + (30 km)^2) = 58.31 km

Answer 2

Ships Distance Calculation

User

two ships leave a harbour. the first leaves at 12:00pm at a speed of 20km/hr on a course 30 degrees east of north. the second leaves at 12:30pm at a speed of 15km/hr on a course of 20 degrees east of south. at 2:30pm, what is the distance between the ships? use the distance, speed, time triangle in your answer, along with basic eleventh grade trigonometric ratios.

ChatGPT

To find the distance between the two ships at 2:30 pm, we need to calculate their respective positions at that time and then use the distance, speed, time triangle along with trigonometric ratios to determine the distance between them.

First, let's calculate the positions of both ships at 2:30 pm.

Ship 1:

Time elapsed since 12:00 pm = 2.5 hours

Speed of ship 1 = 20 km/hr

Distance traveled by ship 1 = Speed × Time = 20 km/hr × 2.5 hrs = 50 km

The direction of ship 1's course is 30 degrees east of north. We can represent this as a vector:

Ship 1's North Component = 50 km × sin(30 degrees)

Ship 1's East Component = 50 km × cos(30 degrees)

Ship 1's North Component = 50 km × 0.5 = 25 km

Ship 1's East Component = 50 km × √(3)/2 ≈ 43.30 km

So, at 2:30 pm, Ship 1 is approximately 25 km north and 43.30 km east of the starting point.

Now let's calculate the position of Ship 2 at 2:30 pm.

Time elapsed since 12:30 pm = 2 hours

Speed of ship 2 = 15 km/hr

Distance traveled by ship 2 = Speed × Time = 15 km/hr × 2 hrs = 30 km

The direction of ship 2's course is 20 degrees east of south. To represent this as a vector, we'll use negative values for the southward direction:

Ship 2's South Component = 30 km × sin(20 degrees)

Ship 2's East Component = 30 km × cos(20 degrees)

Ship 2's South Component = 30 km × 0.3420 ≈ 10.26 km

Ship 2's East Component = 30 km × 0.9397 ≈ 28.19 km

So, at 2:30 pm, Ship 2 is approximately 10.26 km south and 28.19 km east of the starting point.

Now we need to find the distance between the positions of Ship 1 and Ship 2. This can be calculated using the distance, speed, time triangle and trigonometric ratios.

Let D be the distance between the ships. We can use the Pythagorean theorem to find D:

D² = (North Separation)² + (East Separation)²

D² = (25 km - (-10.26 km))² + (43.30 km - 28.19 km)²

D² = (35.26 km)² + (15.11 km)²

D² = 1242.7076 km² + 227.8321 km²

D² = 1470.5397 km²

D ≈ √1470.5397 km ≈ 38.37 km

So, at 2:30 pm, the distance between the two ships is approximately 38.37 kilometers.