t noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 16 knots and ship B is sailing north at 22 knots. How fast (in knots) is the distance between the ships changing at 7 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)
Note: Draw yourself a diagram which shows where the ships are at noon and where they are "some time" later on. You will need to use geometry to work out a formula which tells you how far apart the ships are at time t, and you will need to use "distance = velocity * time" to work out how far the ships have travelled after time t.
The Note says it all. How can I help you.
just help me find out the formula which tells how far apart the ships are
150
To solve this problem, we can use the concept of relative velocity. Let's first draw a diagram to understand the situation:
At noon:
Ship A is 30 nautical miles due west of ship B.
Now, let's introduce some variables:
Let x = distance ship A travels west (in nautical miles) from noon to 7 PM.
Let y = distance ship B travels north (in nautical miles) from noon to 7 PM.
Since ship A is sailing west at 16 knots, and 1 knot is a speed of 1 nautical mile per hour, ship A will travel 16x nautical miles in x hours.
Similarly, since ship B is sailing north at 22 knots, ship B will travel 22y nautical miles in y hours.
Now, we can calculate the position of the ships at 7 PM:
Ship A: 30 + 16x nautical miles due west of the noon position.
Ship B: 22y nautical miles due north of the noon position.
Using the Pythagorean theorem, the distance between the two ships can be calculated as follows:
Distance = √[(30 + 16x)^2 + (22y)^2]
To find how fast the distance between the ships is changing, we can take the derivative of the distance equation with respect to time (t), using the chain rule:
d(Distance)/dt = d(√[(30 + 16x)^2 + (22y)^2])/dt
Now, let's calculate the first derivative (d(√[(30 + 16x)^2 + (22y)^2])/dt):
d(Distance)/dt = (1/2)((30 + 16x)^2 + (22y)^2)^(-1/2) * [2(30 + 16x)(16(dx/dt)) + 2(22y)(dy/dt)]
At 7 PM, t = 7 hours. We need to find dx/dt and dy/dt at t = 7.
Given that x represents the distance ship A travels in hours, and y represents the distance ship B travels in hours, we can write:
x = 7
y = 7
Substituting these values into the first derivative equation, we can calculate the rate of change of distance between the ships at 7 PM:
d(Distance)/dt = (1/2)((30 + 16(7))^2 + (22(7))^2)^(-1/2) * [2(30 + 16(7))(16) + 2(22(7))(22)]
Simplifying this equation will give us the rate of change of distance between the ships at 7 PM:
d(Distance)/dt = (1/2)((30 + 112)^2 + (154)^2)^(-1/2) * [2(30 + 112)(16) + 2(154)(22)]
Evaluating this equation will give us the final answer, which is the speed at which the distance between the ships is changing at 7 PM.