Two ships leave a harbor at the same timeOne ship travels on a bearing s * 14 deg * w at 15 miles per hour. The other ship travels on bearing N * 75 deg * E at 10 miles per hourHow far apart will the ships be after 3 hours? The distance is approximatelymiles(Round to the nearest tenth as needed.)
To solve this problem, we can break down the motion of each ship into its north-south and east-west components.
For the first ship:
- Speed = 15 miles per hour
- Bearing = s * 14 deg * w
North-South Component: 15 * cos(14 deg) = 14.3 miles per hour
East-West Component: 15 * sin(14 deg) = 4.0 miles per hour
For the second ship:
- Speed = 10 miles per hour
- Bearing = N * 75 deg * E
North-South Component: 10 * cos(75 deg) = 2.6 miles per hour
East-West Component: 10 * sin(75 deg) = 9.5 miles per hour
After 3 hours, the ships will be:
- First ship: 3 * 14.3 = 42.9 miles north, 3 * 4.0 = 12 miles west
- Second ship: 3 * 2.6 = 7.8 miles north, 3 * 9.5 = 28.5 miles east
- Distance between the ships = sqrt((42.9 - 7.8)^2 + (12 - 28.5)^2) = sqrt((35.1)^2 + (-16.5)^2) = sqrt(1236.01 + 272.25) = sqrt(1508.26) ≈ 38.8 miles
Therefore, the ships will be approximately 38.8 miles apart after 3 hours.