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At 1 PM, ship A leaves port heading due ...

At 1 PM, ship A leaves port heading due west at x miles per hour. Two hour later, Ship B is 100 miles due south of the same port and heading due north at y miles per hour. At 5 PM, how far apart are the ships?

A

`sqrt((4x)^2 + (100 + 2y)^2)`

B

`x + y`

C

`sqrt(x^2 + y^2)`

D

`sqrt((4x)^2 + (100-2y)^2))`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will analyze the movements of both ships and apply the Pythagorean theorem to find the distance between them at 5 PM. ### Step-by-Step Solution: 1. **Identify the positions of the ships at 5 PM:** - Ship A leaves the port at 1 PM heading due west at a speed of \( x \) miles per hour. By 5 PM, it has been traveling for 4 hours. - Ship B leaves the port at 3 PM heading due north at a speed of \( y \) miles per hour. By 5 PM, it has been traveling for 2 hours. 2. **Calculate the distance traveled by Ship A:** - Distance traveled by Ship A in 4 hours = Speed × Time = \( x \times 4 = 4x \) miles. - Therefore, at 5 PM, Ship A is \( 4x \) miles west of the port. 3. **Calculate the distance traveled by Ship B:** - Distance traveled by Ship B in 2 hours = Speed × Time = \( y \times 2 = 2y \) miles. - Since Ship B started 100 miles south of the port and is heading north, at 5 PM, it will be \( 100 - 2y \) miles south of the port. 4. **Set up the right triangle:** - Let point P be the port, point A be the position of Ship A, and point C be the position of Ship B at 5 PM. - The distance from the port to Ship A (AP) is \( 4x \) miles (horizontal distance). - The distance from the port to Ship B (PC) is \( 100 - 2y \) miles (vertical distance). 5. **Use the Pythagorean theorem to find the distance between the two ships (AC):** - According to the Pythagorean theorem: \[ AC^2 = AP^2 + PC^2 \] - Substituting the distances: \[ AC^2 = (4x)^2 + (100 - 2y)^2 \] - Therefore: \[ AC^2 = 16x^2 + (100 - 2y)^2 \] 6. **Take the square root to find AC:** - To find the distance \( AC \): \[ AC = \sqrt{16x^2 + (100 - 2y)^2} \] ### Final Answer: The distance between Ship A and Ship B at 5 PM is: \[ AC = \sqrt{16x^2 + (100 - 2y)^2} \]
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