At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 22 knots and ship B is sailing north at 25 knots. How fast (in knots) is the distance between the ships changing at 6 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
Let the time passes since noon be t hours
label the position of ship A as P, and that of ship B as Q
clearly PAQ will form a right-angled triangle, with AP=25t nautical miles and BQ = 22t nautical miles
Let the distance PQ = d
d^2 = (25t)^2 + (22t)^2 = 1109t^2
2d dd/dt = 2218t
dd/dt = 2218t/(2d) = 1109/d knots
when time is 6 pm, t = 6
d^2 = 1109(36)
d = 6√1109
dd/dt = 1109/(6√1109) = appr + 5.55 knots
To find the speed at which the distance between the ships is changing, we can use the concept of relative velocity. The relative velocity is the vector difference between the velocities of the two objects.
Let's break down the problem and find the components of velocity for each ship.
- Ship A: sailing west at 22 knots
- The velocity of ship A can be represented as -22i (the negative sign indicates westward direction)
- Ship B: sailing north at 25 knots
- The velocity of ship B can be represented as 25j (the positive sign indicates northward direction)
Now, let's find the velocity of ship A relative to ship B by subtracting the velocity of ship B from the velocity of ship A:
Relative velocity (v_relative) = -22i - 25j
Now, we need to find the magnitude of the relative velocity (|v_relative|) at 6 PM. To do this, we need to know the time elapsed between noon and 6 PM, as well as the distance traveled by each ship during that time.
Given that 6 hours have passed from noon to 6 PM, we can calculate the distance traveled by ship A as follows:
Distance traveled by ship A = 22 knots/hour * 6 hours = 132 nautical miles
Since ship A is 30 nautical miles due west of ship B at noon, after 6 hours, ship A would be 132 nautical miles - 30 nautical miles = 102 nautical miles due west of ship B.
To find the magnitude of the relative velocity, we need to calculate the difference in the x-coordinate (westward) and the y-coordinate (northward) of ship A relative to ship B:
x-component: -102 nautical miles - 0 nautical miles = -102
y-component: 0 nautical miles - 30 nautical miles = -30
Using the Pythagorean theorem, we can find the magnitude of the relative velocity:
|v_relative| = sqrt((-102)^2 + (-30)^2) ≈ 107.61 knots
Therefore, the speed at which the distance between the ships is changing at 6 PM is approximately 107.61 knots.