At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 19 knots and ship B is sailing north at 20 knots. How fast (in knots) is the distance between the ships changing at 7 PM?
i really don't have any idea what to do...
The relative velocity vector is
Vab = Va - Vb = -19*t i + 20t j
where i is a unit vector east and j is a unit vector north.
The magnitude of that vector is the rate of change of the distance between them, and is
|Vab| = sqrt[19^2 + 20^2] = 27.59 knots
That rate is a constant, and does not depend upon time of day nor the initital 40 n.m. separation. You have been given more information than you need.
The actual separation distance at any time DOES depend upon time and initial separation.
To find the rate at which the distance between the ships is changing at 7 PM, we can use the concept of rates of change. Let's break down the problem step by step.
First, let's visualize the positions of the ships at noon:
- Ship A is 40 nautical miles due west of ship B.
Now, let's consider the rates of change of their positions:
- Ship A is sailing west at 19 knots.
- Ship B is sailing north at 20 knots.
To find the speed at which the distance between the ships is changing, we need to consider their relative velocities. Since the ships are moving in different directions, we can use the Pythagorean theorem to find the distance between them as a function of time.
Let's assume t represents the time in hours after noon. We'll consider the horizontal and vertical components separately.
Horizontal Component:
- The horizontal distance of Ship A from its original position is given by dA(t) = 40 - 19t.
- The horizontal distance of Ship B from its original position is given by dB(t) = 0 (as it is moving north).
Vertical Component:
- The vertical distance of Ship A from its original position is given by A(t) = 0 (as it is moving west).
- The vertical distance of Ship B from its original position is given by B(t) = 20t.
Using the Pythagorean theorem, the distance between the ships at time t is given by:
D(t) = sqrt((dA(t))^2 + (dB(t))^2)
= sqrt((40 - 19t)^2 + (20t)^2)
To find the speed at which the distance between the ships is changing at 7 PM, we need to find the derivative of D(t) with respect to time and then evaluate it at t = 7. Let's calculate it:
D'(t) = (dD(t)/dt) = (1/2) * (40 - 19t)^(-1/2) * (-19) * (-1) + 2 * (20t) * (20)
= 19(40 - 19t)^(-1/2) + 800t
Now, substitute t = 7 into the equation to find the speed at 7 PM:
D'(7) = 19(40 - 19(7))^(-1/2) + 800(7)
= 19(40 - 133)^(-1/2) + 5600
= 19(−93)^(-1/2) + 5600
= 19/√93 + 5600
Therefore, the speed at which the distance between the ships is changing at 7 PM is 19/√93 + 5600 knots.