A ship leaves port at 6 am travelling due east at 12mph. Another ship leaves port at 11 am travelling due north at 15 mph. How far apart, to the nearest tenth of a mile are the 2 ships at 11 pm?
by 11 pm, ship1 has gone 17*12 = 204 miles east
Also, ship2 has gone 180 miles north
the final separation is thus
d^2 = 204^2 + 180^2
d = 272.1 miles
To find the distance between the two ships at 11 pm, we need to find their positions at that time.
First, let's determine the position of the eastbound ship at 11 pm. We know it started at 6 am and traveled for 17 hours (11 pm - 6 am = 17 hours). Since it traveled at a speed of 12 mph, the eastbound ship has traveled a distance of 12 mph x 17 hours = 204 miles east.
Next, let's determine the position of the northbound ship at 11 pm. We know it started at 11 am and traveled for 12 hours (11 pm - 11 am = 12 hours). Since it traveled at a speed of 15 mph, the northbound ship has traveled a distance of 15 mph x 12 hours = 180 miles north.
Now, we can calculate the distance between the two ships using the Pythagorean theorem. The distance is the square root of the sum of the squares of the distances traveled in the east and north directions.
Distance = √((204 miles)^2 + (180 miles)^2)
Calculating this expression, we find:
Distance ≈ √(41616 + 32400) ≈ √(74016) ≈ 272.0 miles.
Therefore, to the nearest tenth of a mile, the two ships are approximately 272.0 miles apart at 11 pm.