At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 24 knots and ship B is sailing north at 22 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.)
did you make a diagram,
that is essential in these type of questions.
let the time be t hours after noon.
In my diagram I have a right-angled triangle with the base as (40 + 24t) and a height of 22t
let s be the distance between them
s^2 = (40+24t)^2 + (22t)^2 (equation #1)
then 2s(ds/dt) = 2(24)(40+24t) + 2(22)(22t) (equation #2)
use t=4 in #1 to find s, then use
that s and t=4 in #2
To solve this problem, we can use the concept of rates of change in distance and Pythagoras' theorem.
Step 1: Determine the positions of the ships at 6 PM.
Since ship A is sailing west at 24 knots, it will have traveled 24 * 6 = 144 nautical miles by 6 PM. Therefore, ship A will be 40 - 144 = -104 nautical miles west of its original position at noon. Similarly, since ship B is sailing north at 22 knots, it will have traveled 22 * 6 = 132 nautical miles by 6 PM. Therefore, ship B will be 0 + 132 = 132 nautical miles north of its original position at noon.
Step 2: Calculate the distance between the ships at 6 PM.
Using Pythagoras' theorem, we can calculate the distance between the ships at 6 PM. The distance can be found using the formula: distance = sqrt((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) and (x2, y2) are the coordinates of the ships at noon and 6 PM respectively. In this case, (x1, y1) = (0, 0) and (x2, y2) = (-104, 132).
Therefore, the distance between the ships at 6 PM is: distance = sqrt((-104 - 0)^2 + (132 - 0)^2) = sqrt(10816 + 17424) = sqrt(28240) ≈ 168.1 nautical miles.
Step 3: Calculate the rate of change of the distance between the ships at 6 PM.
To find the rate of change of the distance between the ships at 6 PM, we can differentiate the distance formula with respect to time. The rate of change of the distance between the ships is given by: rate = (dx/dt * (x2 - x1) + dy/dt * (y2 - y1)) / distance, where dx/dt and dy/dt are the rates of change of the x and y coordinates respectively.
In this case, dx/dt (the rate at which ship A is moving west) is -24 knots, dy/dt (the rate at which ship B is moving north) is 22 knots, (x2 - x1) = -104 - 0 = -104 nautical miles, (y2 - y1) = 132 - 0 = 132 nautical miles, and the distance is 168.1 nautical miles.
Therefore, the rate of change of the distance between the ships at 6 PM is: rate = (-24 * -104 + 22 * 132) / 168.1 = (2496 + 2904) / 168.1 = 5399 / 168.1 ≈ 32.1 knots.
So, the distance between the ships is changing at a rate of approximately 32.1 knots at 6 PM.
To find the speed at which the distance between the ships is changing, we can use the concept of relative velocity.
Step 1: Calculate the positions of the ships at 6 PM.
Ship A has been sailing west at a speed of 24 knots for 6 hours. So, the position of Ship A at 6 PM will be 40 nautical miles + (24 knots * 6 hours) due west of its original position, which is 40 + (24 * 6) = 184 nautical miles.
Ship B has been sailing north at a speed of 22 knots for 6 hours. So, the position of Ship B at 6 PM will be 22 knots * 6 hours due north of its original position, which is 22 * 6 = 132 nautical miles.
Step 2: Calculate the distance between the ships at 6 PM.
To find the distance between the two ships, we can use the Pythagorean theorem. The distance between the two ships is the hypotenuse of a right triangle formed by the positions of the ships.
Using the distance formula: distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
where (x1, y1) are the coordinates of Ship A and (x2, y2) are the coordinates of Ship B.
Distance = sqrt((184 - 0)^2 + (0 - 132)^2) = sqrt(33856 + 17424) = sqrt(51280) ≈ 226.43 nautical miles
Step 3: Calculate the rate at which the distance between the ships is changing.
To find how fast the distance is changing, we need to differentiate the distance formula with respect to time.
Differentiating with respect to time (t):
d(distance) / dt = d(sqrt((x2 - x1)^2 + (y2 - y1)^2)) / dt
= (1/2) * (2 * (x2 - x1) * (dx2/dt) + 2 * (y2 - y1) * (dy2/dt)) / sqrt((x2 - x1)^2 + (y2 - y1)^2)
Since the ships are moving horizontally and vertically, dx2/dt and dy2/dt can be represented as the speed of each ship in their respective directions.
dx2/dt = 0 knots/h (since Ship B is moving north, not horizontally)
dy2/dt = 22 knots/h (the speed of Ship B moving north)
With x1 = 0, y1 = 0, x2 = 184, y2 = 132, we can substitute the values into the formula:
d(distance) / dt = (1/2) * (2 * (184 - 0) * 0 + 2 * (132 - 0) * 22) / sqrt((184 - 0)^2 + (132 - 0)^2)
= (1/2) * (2 * 132 * 22) / sqrt(33856 + 17424)
= (1/2) * 5808 / sqrt(51280)
= 2904 / sqrt(51280)
≈ 40.84 knots
Therefore, the speed at which the distance between the ships is changing at 6 PM is approximately 40.84 knots.