a ship leaves a port at 6 am traveling due east at 12mph. another ship leaves a port 11 am traveling due north at 15 mph. how far apart are they at 11pm to the nearest tenth mile?
eastbound ship went for 17 hrs at 12 mph ---> 204 miles
northbound ship went for 12 hrs at 15 mph ---> 180
solve right-angled triangle ,
by finding hypotenuse with legs of 204 and 180
To determine the distance between the two ships at 11 pm, we need to calculate their respective positions at that time.
Let's start by determining the position of the first ship, which left the port at 6 am and is traveling due east at a speed of 12 mph. By 11 pm, this ship has been traveling for 17 hours (11 pm - 6 am). Multiplying the speed of 12 mph by the traveling time of 17 hours, we find that the ship has traveled 204 miles due east.
Now, let's calculate the position of the second ship, which left the port at 11 am and is traveling due north at a speed of 15 mph. By 11 pm, this ship has been traveling for 12 hours (11 pm - 11 am). Multiplying the speed of 15 mph by the traveling time of 12 hours, we find that the ship has traveled 180 miles due north.
To determine the distance between the two ships, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
In this case, the positions of the two ships form the sides of a right triangle. The distance between the two ships is the hypotenuse, and the distances they have traveled (204 miles east and 180 miles north) are the other two sides.
Using the Pythagorean theorem, we can calculate the distance between the two ships:
distance^2 = (204 miles)^2 + (180 miles)^2
distance^2 = 41616 miles^2 + 32400 miles^2
distance^2 = 74016 miles^2
distance ≈ 271.8 miles (rounded to the nearest tenth)
Therefore, the ships are approximately 271.8 miles apart at 11 pm.