At noon, ship A is 50 nautical miles due west of ship B. Ship A is sailing west at 20 knots and ship B is sailing north at 16 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.)
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.
Following your note, I made a diagram and let the time at "some time" later be t hours.
I see a right-angled triangle.
Let the distance between them be D nmiles.
D^2 = (16t)^2 + (50+20t)^2
2D(dD/dt) = 2(16t)(16) + 2(50+20t)(20)
dD/dt = (656t + 1000)/D
At 6:00, t=6
D^2 = 38116
D = 195.2332
dD/dt = (656(6) + 1000)/195.2332
= 25.28 knots
check my arithmetic.
25.28
fools
= 82.52
It's definitely 25.28
To solve this problem, we need to use geometry and the concept of relative velocity.
1. First, let's create a diagram to visualize the situation. Place ship A at the origin (0,0) and ship B at (50,0), where east is the positive x-axis and north is the positive y-axis.
2. Ship A travels west at 20 knots, which means its position can be represented by the equation x = -20t, where t is the time in hours.
3. Ship B travels north at 16 knots, so its position is given by y = 16t.
4. To find the distance between the ships at any given time t, we can use the distance formula: distance = sqrt((x2 - x1)^2 + (y2 - y1)^2). Substituting the positions of the ships, we have distance = sqrt((50 + 20t)^2 + (16t)^2).
5. To find how fast the distance between the ships is changing, we need to differentiate the distance equation with respect to time t.
6. Applying the chain rule, we get:
d(distance)/dt = (1/2) * (2(50 + 20t)(20) + 2(16t)(16))
Simplifying, we have: d(distance)/dt = 20(50 + 20t) + 16(16t)
= 1000 + 400t + 256t
= 656t + 1000
7. Now, let's find the time t when it is 6 PM. Since the given time is in hours from noon, we have t = 6.
8. Substituting t=6 into the equation from step 6, we get:
d(distance)/dt = 656(6) + 1000
= 3936 + 1000
= 4936
Therefore, at 6 PM, the distance between the ships is changing at a rate of 4936 knots per hour.