At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 21 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
this is a cal problem.
i type it in the computer , its says its wrong. but thanks
i think ur answer is wrong
I got the same answer as Damon.
Damon is right.
To find the rate at which the distance between the ships is changing at 4 PM, we can use the concept of relative velocity.
Let's break down the problem and determine the positions of the ships at 4 PM.
Ship A started 12 hours ago and has been sailing west at 21 knots for that entire time. So, ship A would have traveled a distance of 12 hours * 21 knots = 252 nautical miles due west.
Ship B started 4 hours ago and has been sailing north at 15 knots for that entire time. So, ship B would have traveled a distance of 4 hours * 15 knots = 60 nautical miles due north.
Now, we can consider the positions of the ships at 4 PM:
- Ship A is 252 nautical miles due west of its starting point.
- Ship B is 60 nautical miles due north of its starting point.
To find the distance between the ships at 4 PM, we can use the Pythagorean theorem:
Distance^2 = (252 nautical miles)^2 + (60 nautical miles)^2
Calculating this, we get:
Distance^2 = 63504 + 3600
Distance^2 = 67104
Distance = √67104
Distance ≈ 259.14 nautical miles
Now that we know the distance between the ships at 4 PM, we need to find the rate at which this distance is changing. We can use the concept of relative velocity to do this.
The velocity of ship A is given as 21 knots, but we need to consider the velocity in the west direction, which is negative. So, the velocity of ship A would be -21 knots.
The velocity of ship B is given as 15 knots, but we need to consider the velocity in the north direction. So, the velocity of ship B would be 15 knots.
To find the rate of change of the distance between the ships, we can use the formula:
Rate of change of distance = √(velocity of A)^2 + (velocity of B)^2
Rate of change of distance = √((-21 knots)^2 + (15 knots)^2)
Rate of change of distance = √(441 + 225)
Rate of change of distance = √666
Rate of change of distance ≈ 25.81 knots
Therefore, the rate at which the distance between the ships is changing at 4 PM is approximately 25.81 knots.
after 4 hours
x = - 30 - 21*4 = - 114
y = 15*4 = 60
dx/dt = -21
dy/dt = 15
D at 4 hr = sqrt(114^2+60^2) = 129
D^2 = x^2 + y^2
2 D dD/dt = 2 x dx/dt + 2 y dy/dt
129 dD/dt = -114(-21) + 60(15)
dD/dt = 25.5 knots