At noon, ship A is 10 nautical miles due west of ship B. Ship A is sailing west at 15 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 3 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
To solve this problem, we need to use the concept of relative velocity. Let's break it down into steps:
Step 1: Determine the initial distance between the ships
At noon, ship A is 10 nautical miles due west of ship B. We can consider this as the side of a right triangle, with ship A being the base and ship B being the perpendicular side. Using the Pythagorean theorem, we can calculate the initial distance between the ships (d0). Given that ship A is sailing west (leftwards) and ship B is sailing north (upwards), the triangle looks like this:
B
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The length of side A is 10 nautical miles, and the length of side B is 0 nautical miles. Therefore, the initial distance between the ships (d0) can be calculated as follows:
d0 = √(A^2 + B^2) = √(10^2 + 0^2) = √100 = 10 nautical miles
Step 2: Determine the speeds of ship A and ship B
Ship A is sailing west at 15 knots, and ship B is sailing north at 15 knots.
Step 3: Determine the rate of change of distance
To find the rate at which the distance between the ships is changing, we need to differentiate the equation from Step 1 with respect to time (t) and then substitute the values:
d/dt(d0) = d/dt(√(A^2 + B^2))
Note: Since ship A is moving along the x-axis, its position is a function of time: A = -15t
Differentiating with respect to time:
d/dt(d0) = d/dt(√((-15t)^2 + 0^2))
At 3 PM, t = 3 hours.
Step 4: Calculate the rate of change of distance at 3 PM
Substitute the values into the equation:
d/dt(d0) = d/dt(√((-15t)^2 + 0^2))
= d/dt(√(225t^2 + 0))
= d/dt(√(225t^2))
= (1/2) * (225t^2)^(-1/2) * (450t)
Substitute t = 3 into the equation:
d/dt(d0) = (1/2) * (225 * 3^2)^(-1/2) * (450 * 3)
= (1/2) * (225 * 9)^(-1/2) * (450 * 3)
= (1/2) * (2025)^(-1/2) * 1350
= (1/2) * (1/√(2025)) * 1350
= (1/2) * (1/45) * 1350
= (1/2) * 30
= 15 knots
Therefore, the rate at which the distance between the ships is changing at 3 PM is 15 knots.
To find the speed at which the distance between the ships is changing, we can use the concept of rates and related rates. Let's break down the problem into steps:
Step 1: Determine the distance between the ships at any given time.
Since ship A was 10 nautical miles due west of ship B initially, the distance between them can be found using the Pythagorean theorem. Let's call the distance between them "d":
d² = (10 + 15t)² + (15t)²
where "t" represents the time that has passed since noon in hours.
Remember that the ships' directions are perpendicular to each other, forming a right triangle. The vertical side represents the distance traveled by ship B in relation to time, while the horizontal side represents the distance traveled by ship A.
Step 2: Differentiate the equation with respect to time (t).
Differentiating both sides of the equation will allow us to determine how the distance between the ships changes concerning time.
2d * dd/dt = 2(10 + 15t) * 15 + 2(15t) * 15
Simplifying this equation gives us:
dd/dt = (15(10 + 15t) + 15(15t)) / sqrt((10 + 15t)² + (15t)²)
Step 3: Substitute t = 3 into the equation (given that we are looking for the rate at 3 PM).
Plugging in t = 3 into the equation, we can solve for dd/dt.
dd/dt = (15(10 + 15(3)) + 15(15(3))) / sqrt((10 + 15(3))² + (15(3))²)
Simplifying this equation gives us:
dd/dt = (15(10 + 45) + 15(45)) / sqrt((10 + 45)² + (45)²)
dd/dt = (15(55) + 15(45)) / sqrt((55)² + (45)²)
dd/dt = (825 + 675) / sqrt(3025 + 2025)
dd/dt = 1500 / sqrt(5050)
Now we can find the approximate value of dd/dt using a calculator.
Therefore, the speed at which the distance between the ships is changing at 3 PM is approximately 10.61 knots.