(1 pt) At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 20 knots and ship B is sailing north at 23 knots. How fast (in knots) is the distance between the ships changing at 5 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
To find the speed at which the distance between the ships is changing, we can use the concept of relative velocity.
Let's assume that at 5 PM, ship A has traveled west for t hours, and ship B has traveled north for t hours. We'll denote the distance between the ships at 5 PM as D(t).
Given:
The speed of ship A (westward) = 20 knots
The speed of ship B (northward) = 23 knots
The initial distance between the ships (at noon) = 50 nautical miles
To find the rate at which the distance between the ships is changing at 5 PM, we need to find dD/dt (the derivative of D(t) with respect to t).
Using the Pythagorean theorem, the distance between the ships at any given time can be calculated as:
D(t) = square root of (x^2 + y^2)
Where:
x = (initial distance + (speed of ship A * t))
y = (speed of ship B * t)
From this equation, we can express D(t) as:
D(t) = square root of ((50 + (20t))^2 + (23t)^2)
To find dD/dt, we differentiate D(t) with respect to t:
dD/dt = (1 / 2) * ((50 + (20t))^2 + (23t)^2)^(-1/2) * (2(50 + (20t))(20) + 2(23t)(23))
Simplifying further:
dD/dt = (20(50 + 20t) + 23t(23)) / (square root of (2500 + 2000t + 400t^2 + 529t^2))
At 5 PM, t = 5:
dD/dt = (20(50 + 20(5)) + 23(5)(23)) / (square root of (2500 + 2000(5) + 400(5^2) + 529(5^2)))
Simplifying further:
dD/dt = (20(150) + 23(5)(23)) / (square root of (2500 + 10000 + 1000 + 13225))
dD/dt = (3000 + 529(5)) / (square root of (24500))
dD/dt = (3000 + 2645) / (square root of (24500))
dD/dt = 5645 / (square root of (24500))
Calculating the value:
dD/dt ≈ 380.2 knots per hour
Therefore, the speed at which the distance between the ships is changing at 5 PM is approximately 380.2 knots per hour.
To find the speed at which the distance between the ships is changing, we can use the concept of rates of change.
Let's break down the problem step by step:
Step 1: Determine the position of the ships at 5 PM.
At noon, ship A is 50 nautical miles due west of ship B. Since ship A is sailing west at a constant speed of 20 knots (20 nautical miles per hour), after 5 hours, ship A will have traveled 20 knots/hour * 5 hours = 100 nautical miles to the west. Therefore, at 5 PM, ship A will be 50 nautical miles + 100 nautical miles = 150 nautical miles due west of ship B.
Meanwhile, ship B is sailing north at a speed of 23 knots (23 nautical miles per hour). After 5 hours, ship B will have traveled 23 knots/hour * 5 hours = 115 nautical miles to the north.
So at 5 PM, the position of ship B would be the same, directly under its original position. Ship A, on the other hand, would be 150 nautical miles west of ship B.
Step 2: Calculate the distance between the ships at 5 PM.
To find the distance between the ships, we can use the Pythagorean theorem, as the ships' paths form a right triangle.
The distance between the ships can be calculated using the formula:
distance^2 = (difference in x-coordinates)^2 + (difference in y-coordinates)^2
In this case, since ship B remains stationary in the x-coordinate (east-west direction), the difference in x-coordinates is 150 nautical miles.
Since ship A is located to the west of ship B, the difference in y-coordinates is 0 nautical miles.
Therefore, the distance^2 = (150)^2 + (0)^2 = 22500.
Taking the square root of both sides, the distance between the ships is √22500, which is equal to 150 nautical miles.
Step 3: Find the rate at which the distance between the ships is changing.
We are looking for the speed at which the distance between the ships is changing at 5 PM. In other words, we need to find the derivative of the distance between the ships with respect to time.
Since the distance between the ships is constant (150 nautical miles), the derivative is zero. This means that the speed at which the distance between the ships is changing at 5 PM is 0 knots. Therefore, the distance between the ships is not changing.
In conclusion, the speed at which the distance between the ships is changing at 5 PM is 0 knots.