One ship is sailing south at a rate of 5 knots, and another is sailing east at a rate of 10 knots. At 2 P.M. the second ship was at the place occupied by the first ship one hour before. At what time does the distance between the ships not changing?

Why did it become 5+5t?

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Well, it seems these ships are playing a little game of hide-and-seek on the high seas! To figure out when the distance between them is not changing, we need to find the point where their paths intersect.

The first ship is sailing south at 5 knots, and the second ship is sailing east at 10 knots. So, we can think of this as a right-angled triangle, with the distance between the ships as the hypotenuse.

Now, the distance covered by the first ship in one hour (from 1 P.M. to 2 P.M.) is given by its speed of 5 knots. That means, at 2 P.M., the second ship was at the same position, one hour behind.

To find out when the distance between the ships is not changing, we need to calculate the time it takes for the second ship to catch up to the position of the first ship. Since the second ship is moving east at a rate of 10 knots and the first ship was 5 knots to the south at 2 P.M., we can use the Pythagorean theorem to solve for the time it takes for them to meet.

Using the formula distance equals speed multiplied by time, we can set up the equation:

(10t)^2 = (5t + 5)^2

Simplifying this equation will give us the time it takes for the distance between them to stop changing. But before I do that, let me tell you a ship joke to lighten the mood:

Why don't ships like to go to school?

Because they already know their way around!

Now, back to the problem. Solving the equation will give us:

100t^2 = 25t^2 + 50t + 25

Combining like terms, we get:

75t^2 - 50t - 25 = 0

Using the quadratic formula, we can find the value(s) of t, which will tell us at what time the distance between the ships stops changing.

t = (-(-50) ± √((-50)^2 - 4(75)(-25))) / (2(75))

Calculating the values gives us two solutions, approximately t = 0.43 and t = -0.19. Since time cannot be negative, we discard the negative solution.

Therefore, the time at which the distance between the ships stops changing is approximately 0.43 hours after 2 P.M. or 2:26 P.M.

Hope that clears things up! Remember, ships may sail away, but the laughter will always stay on board!

To find the time at which the distance between the ships is not changing, we need to determine when they will meet or intersect.

Let's start by visualizing their paths. The first ship is sailing south at a rate of 5 knots, and the second ship is sailing east at a rate of 10 knots.

We know that the second ship was at the place occupied by the first ship one hour before they met. So, let's assume they meet at time t.

At time t, the first ship will have been sailing for t hours, which means it has traveled 5t knots southward.

Similarly, at time t, the second ship will have been sailing for t-1 hours, as it was at the place occupied by the first ship one hour earlier. This means it has traveled 10(t-1) knots eastward.

Now, if the ships meet, it means they will be at the same coordinates. Therefore, we can equate their positions:

5t (southward distance traveled by the first ship) = 10(t-1) (eastward distance traveled by the second ship)

Next, let's solve this equation to find the value of t:

5t = 10t - 10
10 = 10t - 5t
10 = 5t
t = 2

So, the ships will meet at time t = 2 hours.

Now, to answer the original question, we want to find the time when the distance between the ships is not changing. This occurs when the ships are at the same coordinates, meaning they have met.

Since they meet at time t = 2 hours, the time at which the distance between the ships is not changing is 2 P.M.

So at 2:00 pm the first ship would be 5 nautical miles directly south of the second ship.

Let the time passes since 2:00 pm be t hours

So the two ships form a right-angled triangle, with a horizontal side of 10t nmiles and a vertical of (5+5t) nmiles
Let the distance between them, the hypotenuse , be D nmiles
D^2 = (10t)^2 + (5+5t)^2
when the distance D is not changing, dD/dt = 0
2D dD/dt = 2(10t)(10) + 2(5+5t)(5)
200t + 50 + 50t = 0
250t = -50
t = -50/250 = -1/5 or -12 minutes

so the distance did not change at 1:48 pm